MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 12

An electron is projected into a magnetic field of

$B = 5\times 10^{-3} T$ and rotates in a circle of radius of

$R = 3\ mm$ . Find the work done by the force due to magnetic field.

0%
$0\ J$
0%
$15\ mJ$
0%
$14\ mJ$
0%
$20\ mJ$

$\mathrm{A}$ magnetic field

$\vec{\mathrm{B}}=\mathrm{B}_{0}\hat{\mathrm{j}}$ exists in the region a

$<\mathrm{x}<2\mathrm{a}$ and

$\vec{\mathrm{B}}=-\mathrm{B}_{0}\hat{\mathrm{j}}$ , in the region

$2\mathrm{a}<\mathrm{x}<3\mathrm{a}$ , where

$\mathrm{B}_{0}$ is a positive constant.

$\mathrm{A}$ positive point charge moving with a velocity

$\vec{\mathrm{v}}=\mathrm{v}_{0}\hat{\mathrm{i}}$ , where

$\mathrm{v}_{0}$ is a positive constant, enters the magnetic field at

$\mathrm{x}=\mathrm{a}$ . The trajectory of the charge in this region can be like,

Explanation

$F=q vB$

Now

$\bar v = v_0 \hat i , \bar B = B_0 \hat j Thus, \bar F = v_0B_0 \hat k$

Thus the particle will start straying towards +z direction.

After it reaches x=2a,

$\bar F = - v_0B_0 \hat k$

There won't be any jerks as force always acts perpendicular to particle motion, and acceleration is finite.

Now it starts to stray towards -z direction.

A bar magnet of magnetic moment $M$ is divided into ' $n$ ' equal parts by cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength $2T$ and held making an angle $60^{0}$ with the direction of the field. When the magnet is released, the kinetic energy of the magnet in the equilibrium position is:

0%
$\dfrac{M}{n}$ J
0%
$Mn$ J
0%
$\dfrac{M}{n^{2}}$ J
0%
$Mn^{2}$ J

Explanation

Answer is A
Net magnetic moment after cutting $= \dfrac{M}{n}$
We know that, $|U|= MB \cos \theta$
In equilibrium position, all the potential energy will be converted into kinetic energy i.e. $KE=|U|$
Given $M_{new}=\dfrac{M}{n}, B=2T, \theta=60^{\circ}$
Therefore, $K.E= \dfrac{M}n\times 2\times \cos 60^{\circ}=\dfrac{M}{n} J$

A charged particle q is moving with a velocity

$\vec{v_{1}}=2\hat{i}m/s$ at a point in a magnetic field

$\vec{B}$ and experiences a force

$\vec{F_{1}}=q(\hat{k}-2\hat{j})N.$ If the same charge moves with velocity

$\vec{v_{2}}=2\hat{j}m/s$ from the same point in that magnetic field and experiences a force

$\vec{F_{2}}=q(2\hat{i}+\hat{k})N$ , the magnetic induction at that point will be :

0%
$\hat{i}+\dfrac{1}{2}\hat{j}-\dfrac{1}{2}\hat{k}$
0%
$-\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}+\hat{k}$
0%
$\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}+\hat{k}$
0%
$-\dfrac{1}{2}\hat{i}+\hat{j}+\dfrac{1}{2}\hat{k}$

Explanation

Let the magnetic induction at required point be

$\bar { B } ={ B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k }$

Magnetic force

$\left( \bar { F } \right)$ on particle of charge (q), moving with velocity

$\bar { V }$ in the magnetic field region of intensity

$\left( \bar { B } \right)$ is given as

$\bar { F } =q\left( \bar { V } \times \bar { B } \right)$

Case 1:

$\bar { { F }_{ 1 } } =q\left( \bar { { V }_{ 1 } } \times \bar { B } \right)$

$q\left( \hat { k } -2\hat { j } \right) =q\left( 2\hat { i } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right)$

$\hat { k } -2\hat { j } =2{ B }_{ y }\hat { k } -2{ B }_{ z }\hat { j }$

$\therefore 2{ B }_{ y }=1,\quad { B }_{ y }={ 1 }/{ 2 }.....\left( i \right)$

$-2=-2{ B }_{ z },\therefore { B }_{ z }=1.....\left( ii \right)$

Case 2:

$\bar { { F }_{ 2 } } =q\left( \bar { { V }_{ 2 } } \times \bar { B } \right)$

$q\left( 2\hat { i } +\hat { k } \right) =q\left( 2\hat { j } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right)$

$2\hat { i } +\hat { k } =-2{ B }_{ x }\hat { k } +2{ B }_{ z }\hat { i }$

$\therefore -2{ B }_{ x }=1,\quad \therefore +{ B }_{ x }={ 1 }/{ 2 }......\left( iii \right)$

Thus from

$\left( i \right) ,\left( ii \right) \& \left( iii \right)$

$\bar { B } =-\cfrac { 1 }{ 2 } \hat { i } +\cfrac { 1 }{ 2 } \hat { j } +\hat { k }$

The magnitude of force per unit length on a wire carrying current

$I$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to

$L$ and current in them is

$I$ ) is:

0%
$\dfrac{+\mu_{0}I^{2}}{\pi L}\hat{j}$
0%
$\dfrac{-\mu_{0}I^{2}}{L}\hat{j}$
0%
$\dfrac{\mu_{0}I^{2}}{2L}\hat{j}$
0%
$\dfrac{\mu_{0}I^{2}}{L}\hat{j}$

Explanation

Net Magnetic field at point O due to both the wires shall be

$\displaystyle B=-2\int_{90^{0}}^{0^{0}}\frac{\mu_0 I}{2\pi l} \sin\theta d\theta$ along

$-k$

$\displaystyle \therefore B= -\dfrac{\mu_0 I}{\pi l}$ along

$-k$

Now,

$F=I(\vec{dl} \times \vec{B})$
Where,

$\vec{dl}$ is along

$i$
Therefore,

$\dfrac{F}{dl}= \dfrac{\mu I^{2}}{\pi l}$ along

$j$

The magnetic field due to a current carrying square loop of side a at a point located symmetrically at a distance of

$\mathrm{a}/2$ from its centre (as shown in figure):

0%
$\displaystyle \dfrac{\sqrt{2}\mu_{0}\mathrm{i}}{\sqrt{3}\pi \mathrm{a}}$
0%
$\displaystyle \dfrac{\mu_{0}\mathrm{i}}{\sqrt{6}\pi \mathrm{a}}$
0%
$\displaystyle \dfrac{2\mu_{0}\mathrm{i}}{\sqrt{3}\pi \mathrm{a}}$
0%
$zero$

A planar coil of area 7

$\mathrm{m}^{2}$ carrying an anti-clockwise current 2 A is placed in an extemal magnetic field

$\vec{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})$ , such that the normal to the plane is along the line

$(3\hat{i}-5\hat{j}+4\hat{k})$ . Select correct statements from the following . ( Consider Normal of the coil and Magnetic moment vectors to be in the same direction )

0%
The potential energy of the coil in the given orientation is $6.4 J$
0%
The angle between the normal (positive) to the coil and the external magnetic field is $\cos^{-1}\ (0.57)$
0%
The potential energy of the coil in the given orientatlon is $3.2 J$
0%
The magnitude of magnetic moment of the coil is about 14 $A{m}^{2}$

Explanation

Area

$A=7m^{2},$ Normal

$\hat{n}=\dfrac{1}{\sqrt{50}}(3\hat{i}-5\hat{j}+4\hat{k})$

$\simeq \dfrac{1}{7}(3\hat{i}-5\hat{j}+4\hat{k})$

$\therefore$ area

$\vec{A}=(3\hat{i}-5\hat{j}+4\hat{k})m^{2}$
Magnetic field

$\vec{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})T$
Current

$I=2A$

$\therefore$ magnetic moment

$\vec{M}=I\vec{A}=(6\hat{i}-10\hat{j}+8\hat{k})Am^{2}$

$|\vec{M}|\simeq 14Am^{2}(2\times 7=14)$
Potential energy U

$=-\vec{M}.\vec{B}$

$=-[1.2-2-2.4]=3.2J$
Let

$\theta$ be the angle between

$\hat{n}$ and

$\vec{B}$ (i.e.,)between

$\vec{M}$ and

$\vec{B}.$
Now,U

$=-MBcos\theta .$

$\therefore cos\theta =-\dfrac{U}{MB}=-\dfrac{3.2}{14\times 0.4}(\because \sqrt{0.17}\simeq 0.4)$

$=-\dfrac{4}{7}=-0.57$

$\theta =\pi -cos^{-1}(0.57)$

A charged particle having charge q experiences a force

$\vec{F}=q(-\vec{j}+\vec{k})N$ in a magnetic field B when it has a velocity

$v_{1} = 1\hat i \ m/s$ . The force becomes

$\vec{F}=q(\vec{i}-\vec{k})N$ when the velocity is changed to

$v_{2}=1\hat{j}m/sec$ . The magnetic induction vector at that point is :

0%
$(\hat{i}+\hat{j}+\hat{k})\ T$
0%
$(\hat{i}-\hat{j}-\hat{k})\ T$
0%
$(-\hat{i}-\hat{j}+\hat{k})\ T$
0%
$(\hat{i}+\hat{j}-\hat{k})\ T$

Explanation

Let the magnetic induction at required point be

$\bar { B } =\left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) T$

Magnetic Force

$\left( \bar { F } \right)$ on charged particle (q) moving with Velocity (V) in magnetic field

$\left( \bar { B } \right)$ is given as

$\bar { F } =q\left( \bar { V } \times \bar { B } \right)$

Case 1:

$q\left( -\hat { j } +\hat { k } \right) =q\left( \hat { i } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right)$

$-\hat { j } +\hat { k } ={ B }_{ y }\hat { k } -{ B }_{ z }\hat { j }$

${ B }_{ z }=1....\left( i \right)$

${ B }_{ y }=+1....\left( ii \right)$

Case 2:

$q\left( \hat { j } -\hat { k } \right) =q\left( \hat { j } \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right)$

$\therefore \hat { j } -\hat { k } =-{ B }_{ x }\hat { k } +{ B }_{ z }\hat { i }$

$\therefore { B }_{ x }=1....\left( iii \right)$

From

$\left( i \right) ,\left( ii \right) \& \left( iii \right)$

$\bar { B } =\left( \hat { i } +\hat { j } +\hat { k } \right) T$ Thus, option A

A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.

0%
2.5 $\mu$ T
0%
5.0 $\mu$ T
0%
2.0 $\mu$ T
0%
None

Explanation

$\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { q\overrightarrow { v } \times \overrightarrow { r } }{ { r }^{ 3 } } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { qvsin\theta }{ { r }^{ 2 } }$

$\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { 2\times 100\times sin{ 30 }^{ 0 } }{ 4 }$

${ 10 }^{ -7 }\times 25$

$\alpha$ particle is moving along a circle of radius

$R$ with a constant angular velocity

$\omega.$ Point

$A$ lies in the same plane at a distance

$2R$ from the centre. Point

$A$ records magnetic field produced by

$\alpha$ particle. If the minimum time interval between two successive times at which

$A$ records zero magnetic field is

$'t'$ , the angular speed

$\omega ,$ in terms of

$t$ is

0%
$\displaystyle \frac{2\pi }{t}$
0%
$\displaystyle \frac{2\pi }{3t}$
0%
$\displaystyle \frac{\pi }{3t}$
0%
$\displaystyle \frac{\pi }{t}$

Explanation

Magnetic field using Biot Savart law:

$\vec{B}= \dfrac{\mu_0}{4\pi}i\dfrac{(\vec{dl}\times \vec{r})}{\vec{r}^3}$

Though the expression above has current i, it can be recast into

$\vec {B}= \dfrac{\mu_0}{4\pi}q\dfrac{(\vec{v}\times \vec{r})}{\vec{r}^3}$

Point A shall record zero magnetic field (due to

$\alpha$ particle) when
the -particle is at position P and Q as shown in figure since

$\vec{v} \parallel \vec{r}$

For any other position of the

$\alpha$ particle

$\vec{v}$ is not parallel

$\vec{r}$

$\therefore \vec{v}\times \vec{r}$ is not equal to zero, magnetic field is not equal to zero.

$\angle POQ =120^0$

$\therefore \omega t= 120^0=\dfrac{2\pi}{3}$

$\Rightarrow \omega = \dfrac{2 \pi}{3t}$

An electric motor is a device to convert electrical energy into mechanical energy. The motor shown below has a rectangular coil

$(15\ cm \times 10\ cm)$ with 100 turns placed in a uniform magnetic fleld

$B=2.5\ T$ . When a current is passed through the coil, lt completes

$50$ revolutions in one second. the power output of the motor is

$1.5\ kW$ . The current rating of the coil should be

0%
1 A
0%
2 A
0%
3 A
0%
4 A

Explanation

Let

$I$ be the current through the coil of 100 turns. Let the magnetic moment

$\vec{M}$ make an angle

$\theta$ at an instant of time.
Then, torque on the coil

$|\vec{\tau }|=MB\sin\theta$
Work done by the torque in each half revolution

$=\dfrac{W}{2}$ , where

$W$ is the work done in one full revolution.

$\therefore \dfrac{W}{2}=\int_{0}^{\pi }MBsin\theta d\theta =IANB\int_{0}^{\pi }\sin\theta d\theta$

$=2IANB$

$\therefore$ work output per revolution

$=W=4INAB$
Hence work output per second

$=Power=4INABf$

$\therefore 1.5\times 10^{3}=4INABf$

$I=\dfrac{1.5\times 10^{3}}{4\times 100\times 150\times 10^{-4}\times 2.5\times 50}$

$\therefore I=\dfrac{15\times 10^{2}}{15\times 50}=2\ A$

A thin rod of mass

$m$ and length

$l$ has charge

$Q$ , distributed uniformly along its length. It is rotated with angular speed

$\omega$ in horizontal plane about an axis, passing through the mid-point and perpendicular to its length. If in the presence of an external magnetic field

$B_{o}$ oriented along the axis of rotation, the total kinetic energy of the rod is reduced to zero, then the angular speed of the rod must be :

0%
$\dfrac{QB_{o}}{2m}$
0%
$\dfrac{2QB_{o}}{m}$
0%
$\dfrac{QB_{o}}{m}$
0%
$\dfrac{QB_{o}}{4m}$

When a particle of charge q is projected with uniform speed u along x - axis of the Cartesian coordinate system

$\left ( \hat{i},\hat{j},\hat{k} \right )$ in the presence of a magnetic field of induction

$\vec{B}$ , the force on q is

$\vec{F}=\frac{qu\vec{B}}{2}\hat{j}$ and when the particle is projected along y - axis with same speed, the force

$\vec{F}=\frac{qu\vec{B}}{2}\left ( -\hat{i}+\sqrt{3}\hat{k} \right )$ . The magnetic field

$\vec{B}$ is

0%
$\frac{B}{2}(\sqrt{3}\hat{i}+\hat{k})$
0%
$\frac{B}{2}(\sqrt{3}\hat{i}-\hat{k})$
0%
$-\frac{B}{2}(\sqrt{3}\hat{i}+\hat{k})$
0%
$\frac{B}{2}(-\sqrt{3}\hat{i}+\hat{k})$

Explanation

Initial velocity along x-axis

$U =U \hat{i}$
Force when projected along x-axis

$F=q\left ( \vec{V}\times \vec{B} \right )$
Let the magnetic field be

$\vec{B}=B\left ( a\hat{i}+b\hat{j}+c\hat{v} \right )$

$V_{1}=U\hat{i}$

$F_{1}=\frac{qUB}{2}\hat{j}=qB\left ( U\hat{i}\times \left ( a\hat{i}+b\hat{j} +c\hat{k}\right ) \right )$

$\frac{qUB}{2}\hat{j}=qBu\left ( v\hat{k}-c\hat{j} \right )$

$b\hat{k}-c\hat{j}=\frac{\hat j}{2}$

$c=-\frac{1}{2},b=0$

$F_{2}=\frac{qU\vec{B}}{2}\left ( -\hat{i} +\sqrt{3}\hat{k}\right ),V_{2}=U\hat{j}$

$F_{2}=q\left ( U\hat{j} \times \vec{B}\left ( a\hat{i}+c\hat{k} \right )\right )$

$\frac{qV\vec{B}}{2}\left ( -\hat{i}+\sqrt{3}\hat{k} \right )=qV\vec{B}\left ( -\hat{k} a+c\hat{i}\right )$

$a=-\frac{\sqrt{3}}{2},c=-\frac{1}{2}$

$\vec{B}=-\vec{B}\left ( \frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2}\hat{k}\right )$

A ring or radius

$R$ moves with a velocity

$v$ in a unifrom static magnetic field

$B$ as shown in diagram. The emf between

$P$ and

$Q$ is

0%
$zero$
0%
$vBR(1-\cos \theta)$
0%
$2vBR\sin \dfrac {\theta}{2}$
0%
$vB \sin\theta$

A charged particle

$A$ of charge

$q = 2\ C$ has velocity

$v = 100 \ m/s$ . When it passes through point

$A$ and has velocity in the direction shown, the strength of magnetic field at point

$B$ due to this moving charge is

$(r = 2\ m)$

0%
$2.5\ \mu T$
0%
$5.0\ \mu T$
0%
$2.0\ \mu T$
0%
$none\ of\ these$

Explanation

Magnetic field using Biot Savart law

$\vec{B}= \dfrac{\mu_0}{4\pi}i\dfrac{(\vec{dl}\times \vec{r})}{\vec{r}^3}$
Though the expression above has current

$i$ , it can be recast into

$\vec {B}= \dfrac{\mu_0}{4\pi}q\dfrac{(\vec{v}\times \vec{r})}{\vec{r}^3}$

$\therefore \left |\vec {B} \right | =( 10^{-7} ) \times 2 \times \dfrac{ 100 \times ( \sin \theta)}{2^2}$

$\therefore \left |\vec {B} \right | =( 10^{-7} ) \times 2 \times \dfrac{ 100 \times \sin 30^o}{4}$

$\therefore \left |\vec {B} \right | = 2.5 \times 10^{-6} \: T = 2.5 \: \mu T$

A charged particle moves through a magnetic field in a direction perpendicular to it. Then the

0%
Speed of the particle remains unchanged
0%
Direction of the particle remains unchanged
0%
Acceleration remains unchanged
0%
Velocity remains unchanged

Explanation

[]We know that

$Force$ on a charged prtice

$q$ due to magnetic field

$\vec{B}$ is given by

$\vec{F}=q(\vec{V}\times\vec{B})$ where

$\vec{V}$ is the velocity of charged particle

Thus,from the expression of force it is clear that

$\vec{F}$ is perpendicular to both

$\vec{V}and \vec{B}$

$Force$ is perpendicular to

$Velocity ,$ this implies

$Speed$ will not be changed only

$Direction$ will be changed

Thus

$Speed$ will remain unchanged

answer A

A positive charge particle with velocity

$\overrightarrow{v} = x\hat{i} + y\hat{j}$ moves in a magnetic field

$\overrightarrow{B} = y\hat{i} + x\hat{j}$ . The magnitude of magnetic force acting on the particle is F. Which one of the following statements are correct ?

0%
no force will act on particle if $x = y$
0%
$F \propto (x^2 - y^2)$ if $x > y$
0%
the force will act along positive Z-axis if $x > y$
0%
the force will act along positive Z-axis if $y > x$

Explanation

The magnetic force on a particle moving with velocity '

$v$ ' in a given magnetic field is given by,

$\overrightarrow {F} = \vec v \times \vec B$

$\overrightarrow {F}= (x\hat{i} + y \hat{j}) \times (y \hat{i} + x \hat{j})$

$\overrightarrow {F}=x^2 \hat{k} + y^2 (-\hat{k})$

$\overrightarrow {F} = x^2 \hat{k} - y^2 \hat{k}$
If

$x = y ,$ then

$F = (x^2 - y^2) \hat{k} = 0$

$F > 0 \quad if \quad x > y$ .
So the force acts in positive

$z\ -$ direction if

$x > y$

A Positive charge particle with velocity

$\vec v=x\hat i+y\hat j$ moves in a magnetic field

$\vec B=y\hat i+x\hat j$ . The magnitude of magnetic force acting on the particles is F. Which one of the following statements are correct :

0%
no force will act on particle of x = y
0%
$F\propto (x^2-y^2)$ if x > y
0%
the force will act along positive Z-axis if x > y
0%
the force will act along positive Z-axis if y > x

Explanation

The magnetic force on a particle moving with velocity 'v' in a given magnetic field is given by

$\vec F =\vec v \times \vec B$

$\vec F= (xi + y j) \times (yi + xj)$

$\vec F = x^2 k + y^2 (-k)$

$\vec F = x^2 k - y^2 k$
If

$x = y , F = (x^2 - y^2) k = 0$

$F > 0$ if

$x > y$ . So the force acts in positive z- direction if x > y

The field created by the current in the loop at point C will be

0%
$-\dfrac{\mu _{0} }{4\pi }\hat{k}$
0%
$-\dfrac{\mu _{0} }{8\pi }\hat{k}$
0%
$-\dfrac{\mu _{0}\sqrt{2} }{\pi }\hat{k}$
0%
none

Explanation

From Fleming's right hand rule , as the current is in anti-clockwise direction , field should be towards the positive z axis, i.e.

$\hat{k}$ .As none of the options are suitable , the answer is none ( Option D ) .

A charged particle is kept at rest in a uniform magnetic field. If the magnetic field increases with time,

0%
the particle starts moving
0%
the particle undergoes circular motion
0%
the particle starts moving in opposite direction
0%
None of the above

The radius of the curved part of the wire is

$R=100\:mm$ , the linear parts of the wire are very long. Find the magnetic induction at the point

$O$ if the wire carrying a current

$I=8.0\:A$ has the shape shown in figure.

0%
$\displaystyle 0.34\:\mu T$
0%
$\displaystyle 0.11\:\mu T$
0%
$\displaystyle 1.1\:\mu T$
0%
$\displaystyle 34\:\mu T$

Explanation

$\vec{B_0}=\vec{B_1}+\vec{B_2}+\vec{B_3}$

$=\dfrac{\mu_0}{4\pi}\dfrac{i}{R}(\vec{-k})+\dfrac{\mu_0}{4\pi}\dfrac{i}{R}\pi(\vec{-i})+\dfrac{\mu_0}{4\pi}\dfrac{i}{R}(\vec{-i})$

$=-\dfrac{\mu_0}{4\pi}\dfrac{i}{R}[\vec{k}+(\pi+1)\vec{i}]$

So,

${\vec{B_0}}=\dfrac{\mu_0}{4\pi}\dfrac{i}{R}\sqrt{1+(\pi+1)^2}$

$=0.34\mu T$

The radius of the curved part of the wire is

$R=100\:mm$ , the linear parts of the wire are very long. Find the magnetic induction at the point

$O$ if the wire carrying a current

$I=8.0\ A$ has the shape shown in figure.

0%
$0.44\:\mu T$
0%
$0.34\:\mu T$
0%
$0.56\:\mu T$
0%
$0.78\:\mu T$

Explanation

Given,

$I=8A$

$R=100mm=0.1m$

$\dfrac{\mu_0}{4\pi}=10^{-7}$

Magnetic field due to semi-circular wire=

$\dfrac{\mu_0 i}{4R}(-\hat{i})$

Magnetic field due to semi-infinite wire along x axis=

$\dfrac{\mu_0 i}{4\pi R}(-\hat{k})$

Magnetic field due to semi-infinite wire along z axis=

$\dfrac{\mu_0 i}{4\pi R}(-\hat i)$

Thus net magnetic field at O is

$\dfrac{\mu_0 i}{4R}(1+\dfrac{1}{\pi})(-\hat{i})+\dfrac{\mu_0 i}{4\pi R}(-\hat{k})$

$B_0=-\dfrac{\mu_0 I}{4\pi R}(\hat k+(\pi+1) \hat i)$

It's magnitude is,

$|B_0|=\dfrac{\mu_0 I}{4\pi R}.\sqrt{(1)^2+(\pi+1)^2}$

$|B_0|=10^{-7}\times \dfrac{8}{0.1}\sqrt{1+(4.14)^2}$

$|B_0|=0.34\mu T$

The correct option is B.

A charge particle of charge

$q$ is moving with speed

$v$ in a circle of radius

$R$ as shown in figure. Then the magnetic field at a point on axis of circle at a distance

$x$ from centre is :

0%
$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{R^{2}}$
0%
$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{(R^{2}+x^{2})}$
0%
$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{x^{2}}$
0%
$\dfrac{\mu _{0}}{4\pi }\dfrac{qVR}{(R^{2}+x^{2})^{3/2}}$

Explanation

Charge

$q$ moving in a circle gives rise to a current

$I=\dfrac{q}{\dfrac{2\pi R}{v}}=\dfrac{qv}{2\pi R}$
The magnetic field on the axis at a distance

$x$ from the center of a current carrying circular loop of radius

$R$ is:

$B_{axis}=\dfrac{\mu_0IR^2}{2(R^2 + x^2)^\frac{3}{2}}=\dfrac{\mu_0\dfrac{qv}{2\pi R}R^2}{2(R^2 + x^2)^\frac{3}{2}}=\dfrac{\mu_0qvR}{4\pi(R^2 + x^2)^\frac{3}{2}}$

The potential energy for a force field

$\vec {F}$ is given by

$U(x,y)=\sin(x+y)$ . The force acting on the particle of mass

$m$ at

$\left(0,\dfrac {\pi}{4}\right)$ is

0%
$1$
0%
$\sqrt {2}$
0%
$\dfrac {1}{\sqrt {2}}$
0%
$0$

A non conductive insulating ring of mass

$m$ and radius

$R$ , having charge

$Q$ uniformly distributed over it's circumference is hanging by a insulated thread with the help of a small smooth ring (not rigidly fixed with bigger ring). A time varying magnetic field

$B=B_0 \sin \omega t$ is switched at

$t=0$ and ring is released at the same time. The average magnetic moment of Ring in time interval

$0$ is:

0%
$\dfrac {B_0q^2R^2}{2\pi m}$
0%
$\dfrac {B_0q^2R^2}{4\pi m}$
0%
$\dfrac {B_0q^2R^2}{\pi m}$
0%
zero

Explanation

From electromagnetic Induction,

$\dfrac {E}{R'} = \dfrac {-d}{dt}BA$

$\therefore \dfrac {E}{R'} = \dfrac {-d}{dt}B_{o}A\sin wt$

$\therefore \dfrac {E}{R'} = - B_{0}wA\cos wt = 1$

Magnetic moment

$\mu = iA$

$t = 0, \dfrac {E}{R'} = B_{o}wA$

Note that

$I = \dfrac {B_{o}q^{2}}{4\pi^{2} m}$

$\therefore \mu = iA = 2\pi R^{2} \times \dfrac {B_{o}q^{2}}{4\pi^{2}m}$

$\therefore \mu = \dfrac {B_{o}q^{2}R^{2}}{2\pi m}$

The magnetic field

$B$ at the centre of a circular coil of radius

$r$ is

$\pi$ times that due to a long straight wire at a distance

$r$ from it, for equal currents. The following diagram shows three cases. In all cases the circular part has radius r and straight ones are infinitely long. For the same current the field

$B$ at centre

$P$ in cases

$1, 2, 3$ has the ratio :

0%
$\left (-\dfrac {\pi}{2}\right ):\left (\dfrac {\pi}{2}\right ):\left (\dfrac {3\pi}{4}-\dfrac {1}{2}\right )$
0%
$\left (-\dfrac {\pi}{2}+1\right ):\left (\dfrac {\pi}{2}+1\right ):\left (\dfrac {3\pi}{4}+\dfrac {1}{2}\right )$
0%
$\left (-\dfrac {\pi}{2}\right ):\left (\dfrac {\pi}{2}\right ):\left (\dfrac {3\pi}{4}\right )$
0%
$\left (-\dfrac {\pi}{2}-1\right ):\left (\dfrac {\pi}{2}-\dfrac {1}{4}\right ):\left (\dfrac {3\pi}{4}+\frac {1}{2}\right )$

Explanation

Let

$B_i$ be the magnetic field at center

$P$ for diagram

$i$ .
Diagram

$1:$
Both semi-infinite wire will have magnetic field in opposite directions at point

$P,$ hence will cancel each other.

$\vec{B_1}= \dfrac{1}{2} \dfrac{\mu_0 i}{2r}\hat{k}$
Diagram

$2:$
Both semi-infinite wires pass through the point

$P,$ hence, magnetic field due to both at

$P$ is zero.

$\vec{B_2}=\dfrac{1}{2} \dfrac{\mu_0 i}{2r}(-\hat{k})$

Diagram

$3:$
Contribution from left semi-infinite wire towards magnetic field at

$P$ is zero.

$\vec{B_3}=\dfrac{1}{2} \dfrac{\mu_0 i}{2\pi r} (\hat{k}) +\dfrac{3}{4} \frac{\mu_0 i}{2r}(-\hat{k})= \dfrac{\mu_0 i}{4r}\dfrac{2}{\pi}\left(\dfrac{1}{2} -\dfrac{3\pi}{4}\right)\hat{k}$

$\therefore B_1 : B_2 : B_3=-\left(\dfrac{\pi}{2}\right) :\left( \dfrac{\pi}{2} \right) : \left( \dfrac{3\pi}{4}-\dfrac{1}{2}\right)$

L is a circular ring made of a uniform wire. Current enters and leaves the rind through straight conductors which, if produced, would have passed through the centre C of the ring. The magnetic field at C :

0%
due to the straight conductors is zero
0%
due to the loop is zero
0%
due to the loop is proportional to $\theta$
0%
due to the loop is proportional to $(\pi -\theta)$

Explanation

Magnetic field at center of the coil due to incoming current and outgoing current is zero since center O lies on the path of the currents.

Let the current along path be

$i_1$ and current along path ADC be

$i_2$ .

Given

$i= i_1 + i_2$

Length of arc ABC

$l_{ABC} = 2\pi R(\dfrac{2\pi -\theta}{2\pi})$

Length of arc ADC

$l_{ADC} = 2\pi R(\dfrac{\theta}{2\pi})$

$\dfrac{i_1}{i_2} = \dfrac{\theta}{2\pi -\theta}$ since the current will be inversely proportional to resistance and resistance is directly proportional to length.

$\therefore i_1 = i\bigg (\dfrac{\theta}{2\pi}\bigg )$

$\therefore i_2 = i\bigg (\dfrac{2\pi -\theta}{2\pi}\bigg )$

$\vec{B_{ABC}} = \dfrac{\mu_0i_1}{2R}\times\bigg (\dfrac{l_{ABC}}{2\pi R}\bigg )(-\hat{k})$

$\vec{B_{ADC}} = \dfrac{\mu_0i_2}{2R}= \dfrac{\mu_0i_2}{2R}\times \bigg (\dfrac{l_{ADC}}{2\pi R}\bigg )(\hat{k})$

$\vec{B} = \vec{B_{ABC}} + \vec{B_{ADC}} = \dfrac{\mu_0}{4\pi R^2}\times(i_1l_{ABC} - i_2l_{ADC})\hat{k}=0$ since

$i_1l_{ABC} = i_2l_{ADC}$

Find the circulation of the vector

$\vec{B}$ around the square path

$T$ with side

$l$ located as shown in the figure above.

0%
$\displaystyle\oint{\vec{B}dr}=2(1-\mu)Bl\sin{\theta}$
0%
$\displaystyle\oint{\vec{B}dr}=(1+\mu)Bl\sin{\theta}$
0%
$\displaystyle\oint{\vec{B}dr}=2(1-+\mu)Bl\sin{\theta}$
0%
$\displaystyle\oint{\vec{B}dr}=(1-\mu)Bl\sin{\theta}$

Explanation

magnetic field in the medium of magnetic permeability

$(\mu)$ becomes

$\mu \times B_{vacuum}$ .

$\oint B.dr = B \sin\theta +[(- B\cos \theta) x + (-\mu B\cos\theta)(l-x)] + (-\mu B\sin\theta)l +[(B\cos\theta)(l-x) + (B\cos\theta)x]$

$\implies \oint B.dr = (1- \mu)Bl \sin\theta$

The radius of the curved part of the wire is

$R=100\:mm$ , the linear parts of the wire are very long. Find the magnetic induction at the point

$O$ if the wire carrying a current

$I=8.0\ A$ has the shape shown in the figure.

0%
$\displaystyle 0.30\:\mu T$
0%
$\displaystyle 0.60\:\mu T$
0%
$\displaystyle 0\:\mu T$
0%
$\displaystyle 0.70\:\mu T$

Explanation

Given,

$I=8A$

$R=100mm=0.1m$

$\dfrac{\mu_0}{4\pi}=10^{-7}$

Magnetic field due to semi-infinite wire at a distance R=

$\dfrac{\mu_0 i}{4\pi R}$

Magnetic field at the center of semi circular wire=

$\dfrac{\mu_0 i}{4R}$

Thus net magnetic field at point O is

$\dfrac{\mu_0 i}{4\pi R}(-\hat{k})$

$+\dfrac{\mu_0 i}{4R}(-\hat{i})$

$+\dfrac{\mu_0 i}{4\pi R}(-\hat{k})$

Hence magnitude of net magnetic field

$\approx 0.30\times 10^{-6}T$

$B_0=-\dfrac{\mu_0 I}{4\pi R}(2\hat k+\pi \hat i)$

It's magnitude is,

$|B_0|=\dfrac{\mu_0 I}{4\pi R}.\sqrt{(2)^2+(\pi)^2}$

$|B_0|=10^{-7}\times \dfrac{8}{0.1}\sqrt{4+(3.14)^2}$

$|B_0|=0.30\mu T$

The correct option is A.

Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are

$\overrightarrow{v}=3\hat i+4\hat j$ and

$\overrightarrow{a}=2\hat i+x\hat j$ . Select the correct options.

0%
$x=-1.5$
0%
$x=3$
0%
Magnetic field has a component along the z-direction
0%
Kinetic energy of the particle is constant

Explanation

Under the influence of magnetic field only, speed and hence, kinetic energy remains constant.

$\overrightarrow{F} \perp \overrightarrow{v}$

$\overrightarrow{F}\cdot \overrightarrow{v}=0 \Rightarrow m\vec a\cdot \overrightarrow{v}=0$

$\Rightarrow m(2\hat{i}+x\hat{j})\cdot (3\hat{i}+4\hat{j})=0$

$\Rightarrow 6+4x=0\Rightarrow x=-1.5$

Let the magnetic field is

$\overrightarrow{B}=a\hat{i}+b\hat{j}+c\hat{k}$ , then from

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

$\Rightarrow m(2\hat i+x\hat j)=q(3\hat{i}+4\hat {j})\times (a\hat{i}+b\hat{j}+c\hat{k})$

$\Rightarrow 2m\hat{i}-1.5 m\hat{j}=q(3b\hat{k}-3c\hat {j}-4a\hat{k}+4c\hat{i})$

$4c=2m\Rightarrow c\neq 0$

Hence, magnetic field has a component along z-direction.

A particle of charge per unit mass

$\alpha$ is released from origin with velocity

$\vec v=v_0\hat i$ in a magnetic field

$\vec B=-B_0\hat k$ for

$x\leq \dfrac {\sqrt 3}{2}\dfrac {v_0}{B_0\alpha}$ and

$\vec B=0$ for

$x > \dfrac {\sqrt 3}{2}\dfrac {v_0}{B_0\alpha}$ . The x-coordinate of the particle at time

$t\left ( > \dfrac {\pi}{3B_0\alpha}\right )$ would be

0%
$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {\sqrt 3}{2}-v_0\left (t-\frac {\pi}{B_0\alpha}\right )$
0%
$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+v_0\left (t-\frac {\pi}{3B_0\alpha}\right )$
0%
$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {v_0}{2}\left (t-\frac {\pi}{3B_0\alpha}\right )$
0%
$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {v_0t}{2}$

Explanation

$r=\dfrac {mv_0}{B_0q}=\dfrac {v_0}{B_0\alpha}, \dfrac {x}{r}=\dfrac {\sqrt 3}{2}=sin \theta$

$\Rightarrow \theta=60^o$

$t_{OA}=\dfrac {T}{6}=\dfrac {\pi}{3B_0\alpha}$

Therefore, x-coordinate of particle at any time

$t > \dfrac {\pi}{3B_0\alpha}$ will be

$x=\dfrac {\sqrt 3}{2}\dfrac {v_0}{B_0\alpha}+v_0\left (t-\dfrac {\pi}{3B_0\pi}\right )cos 60^o$

$=\dfrac {\sqrt 3}{2}\dfrac {v_0}{B_0\alpha}+\dfrac {v_0}{2}\left (t-\dfrac {\pi}{3B_0\alpha}\right )$

A particle having a mass of 0.5 g carries a charge of

$2.5\times 10^{-8}C$ . The particle is given an initial horizontal velocity of

$6\times 10^4 ms^{-1}$ . To keep the particle moving in a horizontal direction

0%
the magnetic field may be perpendicular to the direction of the velocity
0%
the magnetic field should be along the direction of the velocity
0%
magnetic field should have a minimum value of 3.27 T
0%
no magnetic field is required

Explanation

In the absence of a magnetic field, the particle will experience gravitational force mg. As a result, the particle will not continue moving in the horizontal direction but will described a parabolic path. So, a magnetic field must be present and its direction must be perpendicular to the direction of the velocity. The magnetic force experienced by the particle is given by

$\vec F=q(\vec v\times \vec B)$ . The magnitude of the force is

$F=qvB sin \theta$ . If the particle is to move in the horizontal direction, this force must balance the force of gravity, i.e.,

$mg=qvB sin\theta$
The minimum value of B corresponds to

$sin\theta=1$ or

$\theta =90^o$ .
Thus,

$mg=qvB$

$B=\dfrac {mg}{qv}=\dfrac {0.5\times 10^{-3}\times 9.8}{2.5\times 10^{-8}\times 6\times 10^4}=3.27T$ .

An insulating rod of length

$l$ carries a charge

$q$ distributed uniformly on it. The rod is pivoted at its mid-point and is rotated at a frequency

$f$ about a fixed axis perpendicular to the rod and passing through the pivot. The magnetic moment of the rod system is

0%
$\dfrac {1}{12}\pi q f l^2$
0%
$\pi qfl^2$
0%
$\dfrac {1}{6}\pi q f l^2$
0%
$\dfrac {1}{3}\pi q f l^2$

Explanation

Consider a small element

$dx$ at a distance

$x$ from the pivot.

Charge on element

$dq=\dfrac{q}{l}dx$

The rod sweeps a circle of radius

$\dfrac{l}{2}$

Charge

$dq$ rotating at frequency

$f$ will give rise to current

$di=dq \times f=\dfrac{q}{l}fdx$

Magnetic moment of this element

$dx$ is

$d\mu = di \times A= \dfrac{q}{l}fdx \times \pi x^2$

where

$A$ is the area of the circle of radius

$x$ .

$\therefore \mu = \int_{0}^{l}d\mu = \int_{-l/2}^{l/2} \dfrac{q}{l}f \times \pi x^2 dx=\dfrac{\pi qf}{l}\left(\dfrac{\left(\dfrac{l}{2}\right)^3}{3}-\dfrac{\left(\dfrac{-l}{2}\right)^3}{3}\right)=\dfrac{\pi qfl^2}{12}$

A charged particle

$P$ leaves the origin with speed

$v=v_0$ at some inclination with the

$x-$ axis. There is a uniform magnetic field B along the

$x-$ axis.

$P$ strikes a fixed target

$T$ on the

$x-$ axis for a minimum value of

$B=B_0$ . P will also strike T if

0%
$B=2B_0, v=2v_0$
0%
$B=2B_0, v=v_0$
0%
$B=B_0, v=2v_0$
0%
$B=\dfrac {B_0}{2}, v=2v_0$

Explanation

Let

$d$ be the distance of the target from the point of projection.

$P$ will strike

$T$ if

$d$ is an integral multiple of pitch.

$Pitch = \dfrac{2\pi m}{qB}v \cos \theta$

Given,

$P$ strikes

$T$ if

$v=v_0$ and

$B=B_0$

$\therefore Pitch = \dfrac{2\pi m}{qB_0}v_0 \cos \theta$

Hence, pitch will depend on the ratio

$\dfrac{v_0}{B_0}$ .

For the given options,

(A)

$pitch = \dfrac{2\pi m}{q(2B_0)}2v_0 \cos \theta= \dfrac{2\pi m}{qB_0}v_0 \cos \theta$ .

Thus, pitch remains same, hence P will hit the target.

(B)

$pitch = \dfrac{2\pi m}{q(2B_0)}v_0 \cos \theta= \dfrac{1}{2} \dfrac{2\pi m}{qB_0}v_0 \cos \theta$ .

Thus, pitch becomes

$\dfrac{1}{2}$ of the pitch for min value of

$B$ .

Hence,

$P$ will hit the target.

(D) Pitch will become 4 times the pitch for min value of B, hence, P will not hit the target

A charged particle of specific charge (charge/mass)

$\alpha$ is released from origin at time

$t=0$ with velocity

$\vec v=v_0(\hat i+\hat j)$ in a uniform magnetic field

$\vec B=B_0\hat i$ . Coordinates of the particle at time

$t=\dfrac{\pi}{B_0\alpha}$ are

0%
$\displaystyle\left (\frac {v_0}{2B_0\alpha}, \frac {\sqrt 2v_0}{\alpha B_0}, \frac {-v_0}{B_0\alpha}\right )$
0%
$\displaystyle\left (\frac {-v_0}{2B_0\alpha}, 0, 0\right )$
0%
$\displaystyle\left (0, \frac {2v_0}{B_0\alpha}, \frac {v_0\pi}{2B_0\alpha}\right )$
0%
$\displaystyle\left (\frac {v_0\pi}{B_0\alpha}, 0, \frac {-2v_0}{B_0\alpha}\right )$

Explanation

$\alpha=\dfrac {q}{m}$ , path of the particle will be a helix of time period,

$T=\dfrac {2\pi m}{B_0q}=\dfrac {2\pi}{B_0\alpha}$

The given time

$t=\dfrac {\pi}{B_0\alpha}=\dfrac {T}{2}$

$\therefore$ Coordinates of particle at time

$t=T/2$ would be

$(v_xT/2, 0, -2r)$

Here,

$r=\dfrac {mv_0}{B_0q}=\dfrac {v_0}{B_0\alpha}$

$\therefore$ The coordinate are

$\left (\dfrac {v_0\pi}{B_0\alpha}, 0, \dfrac {-2v_0}{B_0\alpha}\right )$

A long circular tube of length

$10\ m$ and radius

$0.3\ m$ carries a current

$I$ along its curved surface as shown. A wire-loop of resistance

$0.005\ ohm$ and of radius

$0.1\ m$ is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as

$I = I_0\cos (300 t)$ where

$I_0$ is constant, If the magnetic moment of the loop is

$N \mu_0 I_0\sin (300 t)$ then '

$N$ ' is

0%
6
0%
8
0%
7
0%
4

Explanation

$l=10m,{ r }_{ 1 }=0.3m$

$R=0.005\Omega ,{ r }_{ 2 }=0.1m$

$I={ I }_{ 0 }\cos { \left( 300t \right) }$
So we get

$\omega =300$
The magnetic field of the tube,

$B=\cfrac { { \mu }_{ 0 }NI }{ L }$
In the problem

$N=1,\quad L=10m,\quad I={ I }_{ 0 }\cos { \left( 300t \right) }$
So,

$B=\cfrac { { \mu }_{ 0 }{ I }_{ 0 }\cos { \left( 300t \right) } }{ 10 }$
Flux,

$Q=\vec { B } \vec { A } =BA=B\pi { r_2 }^{ 2 }$

$Q=\cfrac { { \mu }_{ 0 }{ I }_{ 0 }\cos { \left( 300t \right) } }{ 10 } \times \pi { \left( 0.1 \right) }^{ 2 }$
Induced emf,

$e=-\cfrac { dQ }{ dt } =300{ \mu }_{ 0 }\pi { I }_{ 0 }\sin { \left( 300t \right) } \times { 10 }^{ -3 }$

$e=0.3{ \mu }_{ 0 }\pi { I }_{ 0 }\sin { \left( 300t \right) }$
The current in the ring

$i=\cfrac { e }{ R }$

$i=\cfrac { 0.3\pi { \mu }_{ 0 }{ I }_{ 0 }\sin { \left( 300t \right) } }{ 0.005 }$
or

$i=60{ \mu }_{ 0 }\pi { I }_{ 0 }\sin { \left( 300t \right) }$
Magnetic moment,

$M=iA=i\pi { r }^{ 2 }$

$M=60{ \mu }_{ 0 }\pi { I }_{ 0 }\sin { \left( 300t \right) } \times \pi { \left( 0.1 \right) }^{ 2 }$

$=0.6{ \mu }_{ 0 }{ \pi }^{ 2 }{ I }_{ 0 }\sin { \left( 300t \right) } A$

$=\left( 0.6\times { \pi }^{ 2 } \right) { \mu }_{ 0 }{ I }_{ 0 }\sin { \left( 300t \right) }$

$=6{ \mu }_{ 0 }{ I }_{ 0 }\sin { \left( 300t \right) }$
So, the value of

$N=6$

An electron is moving along the positive x-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative x-axis. This can be done by applying the magnetic field along :

0%
$y-$ axis
0%
$z-$ axis
0%
$y-$ axis only
0%
$z-$ axis only

Explanation

If we apply the magnetic field either in

$y$ or

$z$ direction, the path will be a circle perpendicular to the magnetic field

$B$ .

$B$ along

$y-$ axis: circle in

$xz$ plane, at some point particle will be moving along

$-ve\quad x-$ axis.

$B$ along

$z-$ axis: circle in

$xy$ plane, at some point particle will be moving along

$-ve\quad x-$ axis.

A charged particle with velocity

$\overrightarrow{v}=x\hat i+y\hat j$ moves in a magnetic field

$\overrightarrow{B}=y\hat i+x\hat j$ . The force acting on the particle has magnitude

$F$ . Which one of the following statements is/are correct?

0%
No force will act on charged particle if $x=y$
0%
If $x > y, F\propto(x^2-y^2)$
0%
If $x > y$ , the force will act along z-axis
0%
If $y > x$ , the force will act along y-axis

Explanation

$\overrightarrow{v} = x\hat{i} + y\hat{j}$

$\overrightarrow{B} = y\hat{i} + x\hat{j}$

$\overrightarrow{f} = q(\overrightarrow{v} \times \overrightarrow{B})= q((x\hat{i} + y\hat{j})\times ( y\hat{i} + x\hat{j}))=q(x^2-y^2)\hat{k}$

$x=y$ ,

$\overrightarrow{f} = q(x^2-y^2) =q(0-0)=0$

$x \gt y$ ,

$\overrightarrow{f} = q(x^2-y^2)\hat{k} \gt 0$

$y \gt x$ ,

$\overrightarrow{f} = q(x^2-y^2)\hat{k} \lt 0$ and it points in the

$-ve\quad z-$ axis.

Two very long straight parallel wires carry steady currents

$i$ and

$2i$ in opposite directions. The distance between the wires is

$d$ . at a certain instant of time, a point charge

$q$ is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity

$\vec { v }$ is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is:

0%
$\cfrac { { \mu }_{ 0 }iqv }{ 2\pi d }$
0%
$\cfrac { { \mu }_{ 0 }iqv }{ \pi d }$
0%
$\cfrac { { 3\mu }_{ 0 }iqv }{ 2\pi d }$
0%
zero

Explanation

Magnetic field at

$P$ is perpendicu;ar to paper inward due to both the wires. Charged particle is also projected in the same direction.

So, force on the charged particle is zero as

$\vec { v } \parallel \vec { B } ;\quad \vec { { F }_{ m } } =q(\vec { v } \times \vec { B } )=0$

Find the magnetic field

$\vec B$ .

0%
$(10^{-3}T)(\hat i+\hat j)$
0%
$(2\times 10^{-3}T)\hat i$
0%
$(10^{-3}T)\hat i$
0%
$(2\times 10^{-3}T)(\hat i_\hat j)$

Explanation

Force on a charged particle

$\vec{F} = q( \vec{v} \times \vec{B})$

Let the magnetic field

$\vec{B} =B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$

$1^{st}\ Case:$

$\vec{F} = (10 \times 10^{-6} \times 10^6)( \cos 45^{\circ}\hat{i} + \sin 45^{\circ} \hat{j}) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$

$\therefore 5 \times \sqrt{2} \times 10^{-3}(-\hat{k} )= \dfrac{10}{\sqrt{2}}(\hat{i} + \hat{j})\times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})= \dfrac{10}{\sqrt{2}}[B_z\hat{i} - B_z \hat{j} +(B_y -B_x)\hat{k} ]$

Equating both LHS and RHS for components

$B_z=0$ ;

$B_y-B_x =-10^{-3} ..................................(1)$

$2^{nd}\ Case:$

$\vec{F} = (10 \times 10^{-6} \times 10^6)(\hat{k} ) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$

$\therefore F_2 \hat{j} =(10 \times 10^{-6} \times 10^6)(B_x\hat{j} -B_y \hat{i})=10 \times(B_x\hat{j} -B_y \hat{i})$

$\Rightarrow F_2 =10B_x$ ;

$B_y=0 ...........................(2)$

From equations

$(1)$ and

$(2)$

$B_y-B_x =-10^{-3}= 0 - B_x =-10^{-3}$

$\Rightarrow B_x= 10^{-3}\:T$

$\Rightarrow \vec{B} = (10^{-3}\:T) \hat{i}$

Find the magnitude of the force

$F_2$

0%
$10^{-2}\ N$
0%
$10^{-3}\ N$
0%
$10^{-4}\ N$
0%
$10^{-5}\ N$

Explanation

Force on a charged particle

$\vec{F} = q( \vec{v} \times \vec{B})$

Let the magnetic field

$\vec{B} =B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$

$1^{st}\ Case:$

$\vec{F}= (10 \times 10^{-6} \times 10^6)( \cos 45^{\circ}\hat{i} + \sin 45^{\circ} \hat{j}) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$

$\therefore 5 \times \sqrt{2} \times 10^{-3}(-\hat{k} )= \dfrac{10}{\sqrt{2}}(\hat{i} + \hat{j})\times (B_x \hat{i} + B_y \hat{j} +B_z \hat{k})= \dfrac{10}{\sqrt{2}}[B_z\hat{i} - B_z \hat{j} +(B_y -B_x)\hat{k} ]$

Equating both LHS and RHS for components

$B_z=0$ ;

$B_y-B_x =-10^{-3} .......................................(1)$

$2^{nd}\ Case:$

$\vec{F} = (10 \times 10^{-6} \times 10^6)(\hat{k} ) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$

$\therefore F_2 \hat{j} =(10 \times 10^{-6} \times 10^6)(B_x\hat{j} -B_y \hat{i})=10 \times(B_x\hat{j} -B_y \hat{i})$

$\Rightarrow F_2 =10B_x$ ;

$B_y=0 .............................(2)$

From

$(1)$ and

$(2)$

$B_y-B_x =-10^{-3}= 0 - B_x =-10^{-3}$

$\Rightarrow B_x= 10^{-3}\:T$

$\Rightarrow \vec{B} = (10^{-3}\:T) \hat{i}$

From eqn

$(2)$

$F_2= 10B_x= 10 \times 10^{-3}= 10^{-2}\:N$

The field B at the centre of a circular coil of radius

$r$ is

$\pi$ times that due to a long straight wire at a distance

$r$ from it for equal currents. The figure shows three cases. In all cases the circular part has radius

$r$ and straight ones are infinitely long. For same current, the field B at the centre P in cases 1, 2, 3 has the ratio:

0%
$\displaystyle \left ( - \frac{\pi}{2} \right ) : \frac{\pi}{2} : \left ( \frac{3 \pi}{4} - \frac{1}{2} \right )$
0%
$\displaystyle \left ( - \frac{\pi}{2} + 1\right ) : \left ( \frac{\pi}{2}+ 1 \right ) : \left ( \frac{3 \pi}{4} - \frac{1}{2} \right )$
0%
$\displaystyle -\frac{\pi}{2} : \frac{\pi }{2} : \frac{3 \pi}{4}$
0%
$\displaystyle \left ( - \frac{\pi}{2} - 1\right ) : \left ( \frac{\pi}{4}+ \frac{1}{4} \right ) : \left ( \frac{3 \pi}{4} + \frac{1}{2} \right )$

A wire is bent in the form of a circular arc with a straight portion

$AB$ . Magnetic induction at

$O$ when current

$I$ is flowing in the wire, is

0%
$\cfrac { { { \mu }_{ 0 } } }{ 2r } \left( \pi -\theta +\tan { \theta } \right)$
0%
$\cfrac { { { \mu }_{ 0 } I} }{ 2\pi r } \left( \pi +\theta -\tan { \theta } \right)$
0%
$\cfrac { { { \mu }_{ 0 }I } }{ 2\pi r } \left( \pi -\theta +\tan { \theta } \right)$
0%
$\cfrac { { { \mu }_{ 0 } I} }{ 2\pi r } \left( -\tan { \theta } +\pi -\theta \right)$

A long wire bent as shown in figure carries current

$I$ . If the radius of the semicircular portion is

$a$ , the magnetic field at center

$C$ is:

0%
$\cfrac { { \mu }_{ 0 }I }{ 4a }$
0%
$\cfrac { { \mu }_{ 0 }I }{ 4\pi a } \sqrt { { \pi }^{ 2 }+4 }$
0%
$\cfrac { { \mu }_{ 0 }I }{ 4a }+\cfrac { { \mu }_{ 0 }I }{ 4\pi a }$
0%
$\cfrac { { \mu }_{ 0 }I }{ 4\pi a } \sqrt { { \pi }^{ 2 }-4 }$

Explanation

Due to semicircular part:

${ B }_{ 1 }=\dfrac{1}{2}(\dfrac{\mu_0I}{2a})$

$\quad { B }_{ 2 }=2 \times ( \dfrac{1}{2} \times ( \dfrac{\mu_0I}{2\pi a}))$
The factor of 2 outside is due to two semi-infinite wires.
Hence, resultant field,

$B=\sqrt { { { B }_{ 1 } }^{ 2 }+{ { B }_{ 2 } }^{ 2 } } =\dfrac{\mu _0I}{2a}\sqrt{( ( \dfrac{1}{2})^2 + (\dfrac{1}{\pi })^2)}=\dfrac{\mu _0I}{4\pi a}\sqrt{4 +\pi ^2}$

Select the correct statement

0%
At $t=\dfrac {T_0}{2}$ , coordinate of charge are $\left (\dfrac {P_0}{2}, 0, -2R_0\right )$
0%
At $t=\dfrac {3T_0}{2}$ , coordinate of charge are $\left (\dfrac {3P_0}{2}, 0, 2R_0\right )$
0%
At $t=\dfrac {T_0}{2}$ , coordinate of charge are $\left (P_0, 0, -2R_0\right )$
0%
At $t=\dfrac {3T_0}{2}$ , coordinate of charge are $\left (3P_0, 0, 2R_0\right )$

Explanation

As the magnetic field is along the x-axis, the magnetic force will be along (-) z-axis at

$t=0$ . So, the particle will move in helical path along (1). At

$t=T_0$ , the direction of field changes, so force becomes along +z-direction, and now the particle will move in helical path along (2). It will be moving along x-axis, so that resulting path will be helical.

$t=\dfrac{T_0}{2}$ , particle will be at

$E_1$ .

x-coordinate

$x=\dfrac{P_0}{2}$ (half of pitch)

y-coordinate

$y=0$ (from fig)

and z-coordinate

$z=-2R_0$ (from fig)

Hence, (a) is correct.

Similarly at

$t=\dfrac{3T_0}{2}$ particle will be at

$E_2$ .

The coordinates are (

$\dfrac{3P_0}{2}$ ,0,

$2R_0$ )

Hence, (b) is correct.

From fig, we can see the distance between two extremes:

$E_1$ and

$E_2$ is

$4R_0$ .

Figure shows three cases: in all cases the circular part has radius

$r$ and straight ones are infinitely long. For the same current the ratio of field

$B$ at center

$P$ in the three cases

${B}_{1}:{B}_{2}:{B}_{3}$ is :

0%
$\left( -\cfrac { \pi }{ 2 } \right) :\left( \cfrac { \pi }{ 2 } \right) :\left( \cfrac { 3\pi }{ 4 } -\cfrac { 1 }{ 2 } \right)$
0%
$\left( -\cfrac { \pi }{ 2 } +1 \right) :\left( \cfrac { \pi }{ 2 } +1 \right) :\left( \cfrac { 3\pi }{ 4 } +\cfrac { 1 }{ 2 } \right)$
0%
$\left( -\cfrac { \pi }{ 2 } \right) :\left( \cfrac { \pi }{ 2 } \right) :\left( \cfrac { 3\pi }{ 4 } \right)$
0%
$\left( -\cfrac { \pi }{ 2 } -1 \right) :\left( \cfrac { \pi }{ 2 } -\cfrac { 1 }{ 4 } \right) :\left( \cfrac { 3\pi }{ 4 } +\cfrac { 1 }{ 2 } \right)$

Explanation

$\vec { { B }_{ 1 } } =\cfrac { { \mu }_{ 0 }I }{ 4r } \bigodot$

$\vec { { B }_{ 2 } } =\cfrac { { \mu }_{ 0 }I }{ 4r } \bigotimes$

$\vec { { B }_{ 3 } } =\cfrac { { \mu }_{ 0 }I }{ 2r } \left( \cfrac { 3 }{ 4 } \right) \bigotimes +\cfrac { { \mu }_{ 0 } I}{ 4\pi r } \bigodot$

$=\cfrac { { \mu }_{ 0 }I }{ 4r } \left[ \cfrac { 3 }{ 2 } \bigotimes +\cfrac { 1 }{ \pi } \bigodot \right]$

$=\cfrac { { \mu }_{ 0 }I }{ 4r } \left[ \cfrac { 3\pi -2 }{ 2\pi } \right] \bigotimes =\cfrac { { \mu }_{ 0 }I }{ 4r } \left( \cfrac { 2 }{ \pi } \right) \left( \cfrac { 3\pi }{ 4 } -\cfrac { 1 }{ 2 } \right) \bigotimes$

${ B }_{ 1 }:{ B }_{ 2 }:{ B }_{ 3 }=-\left( \cfrac { \pi }{ 2 } \right) :\left( \cfrac { \pi }{ 2 } \right) :\left( \cfrac { 3\pi }{ 4 } -\cfrac { 1 }{ 2 } \right)$

An otherwise infinite, straight wire has two concentric loops of radii

$a$ and

$b$ carrying equal currents in opposite directions as shown in figure. The magnetic field at the common center is zero for:

0%
$\cfrac { a }{ b } =\cfrac { \pi -1 }{ \pi }$
0%
$\cfrac { a }{ b } =\cfrac { \pi } { \pi +1 }$
0%
$\cfrac { a }{ b }=\cfrac { \pi -1 }{ \pi +1 }$
0%
$\cfrac { a }{ b }=\cfrac { \pi +1 }{ \pi -1 }$

Explanation

${ B }_{ centre }=0$

$\cfrac { { \mu }_{ 0 }I }{ 4\pi b } \bigodot +\cfrac { { \mu }_{ 0 }I }{ 4\pi b } \bigodot +\cfrac { { \mu }_{ 0 }I }{ 2b } \bigodot +\cfrac { { \mu }_{ 0 }I }{ 2a } \bigotimes =0$

$\Rightarrow \cfrac { { \mu }_{ 0 }I }{ 2\pi b } +\cfrac { { \mu }_{ 0 }I }{ 2b } -\cfrac { { \mu }_{ 0 }I }{ 2a } \Rightarrow \cfrac { 1 }{ 2\pi b } +\cfrac { 1 }{ 2b } =\cfrac { 1 }{ 2a }$

$\Rightarrow \cfrac { a }{ b } =\cfrac { \pi }{ \pi +1 }$

The magnetic field at the center of the circular loop as shown in figure, when a single wire is bent to form a circular loop and also extends to form straight sections, is :

0%
$\cfrac { { { \mu }_{ 0 } I} }{ 2R } \bigodot$
0%
$\cfrac { { { \mu }_{ 0 } I} }{ 2R }\left( 1+\cfrac { 1 }{ \pi \sqrt { 2 } } \right) \bigodot$
0%
$\cfrac { { { \mu }_{ 0 } I} }{ 2R }\left( 1-\cfrac { 1 }{ \pi \sqrt { 2 } } \right) \bigotimes$
0%
$\cfrac { { { \mu }_{ 0 } I} }{ R }\left( 1-\cfrac { 1 }{ \pi \sqrt { 2 } } \right) \bigotimes$

Explanation

Magnetic field due to

$AB$ ,

$\vec { { B }_{ 1 } } =\cfrac { { \mu }_{ 0 }I }{ 4\pi R } \left[ \sin { \cfrac { \pi }{ 2 } } -\sin { \cfrac { \pi }{ 4 } } \right] \bigotimes \quad =\cfrac { { \mu }_{ 0 }I }{ 4\pi R } \left[ 1-\cfrac { 1 }{ \sqrt { 2 } } \right] \bigotimes$
Magnetic field due to circular loop

$\vec { { B }_{ 2 } } =\cfrac { { \mu }_{ 0 }I }{ 2R } \bigodot$
Magnetic field due to straight wire

$BC$ ,

$\vec { { B }_{ 3 } } =\cfrac { { \mu }_{ 0 }I }{ 4\pi R } \left[ \sin { \cfrac { \pi }{ 2 } } +\sin { \cfrac { \pi }{ 4 } } \right] \bigodot =\cfrac { { \mu }_{ 0 }I }{ 4\pi R } \left[ 1+\cfrac { 1 }{ \sqrt { 2 } } \right] \bigodot$

$\vec { B } =\vec { { B }_{ 1 } } +\vec { { B }_{ 2 } } +\vec { { B }_{ 3 } }$

$=\left( \cfrac { { \mu }_{ 0 }I }{ 2R } +\cfrac { { 2\mu }_{ 0 }I }{ 4\pi R } \cfrac { 1 }{ \sqrt { 2 } } \right) \bigodot =\cfrac { { \mu }_{ 0 }I }{ 4\pi R } \left[ 1+\cfrac { 1 }{ \sqrt { 2 } \pi } \right] \bigodot$

Positive point chages

$q=+8.00\mu C$ and

$q'=+3.00\mu C$ are moving relative to an observer at point

$P$ , as shown in the figure. the distance

$d$ is

$0.120$

$m$ . When the two charges are at the locations as shown in the figure, what are the magnitude and direction of the net magnetic field they produce at point

$P$ ?
(Take

$v=4.50\times { 10 }^{ 6 }{ m{ s }^{ -1 }}$ and

$v'=9.00\times { 10 }^{ 6 }{ m{ s }^{ -1 }}$ ).

0%
$4.38\times { 10 }^{ -4 }$ $T$ , into the page
0%
$4.38\times { 10 }^{ -4 }$ $T$ , out of the page
0%
$2.16\times { 10 }^{ -4 }$ $T$ , into the page
0%
$2.9\times { 10 }^{ -9 }$ $T$ , out of the page

Explanation

${ B }_{ total }=B+B'=\cfrac { { \mu }_{ 0 } }{ 4\pi } \left( \cfrac { qv }{ { d }^{ 2 } } +\cfrac { q'v' }{ { d }^{ 2 } } \right)$

$B=\cfrac { { \mu }_{ 0 } }{ 4\pi } \left( \cfrac { (8.0\times { 10 }^{ -6 }C)(4.5\times { 10 }^{ 6 }{ ms }^{ -1 }) }{ { \left( 0.120m \right) }^{ 2 } } +\cfrac { (3.0\times { 10 }^{ -6 }C)(9.0\times { 10 }^{ 6 }{ ms }^{ -1 }) }{ { \left( 0.120m \right) }^{ 2 } } \right)$

$B=4.38\times { 10 }^{ -4 }T$
Direction is into the page.

Figure shows an Amperian path

$ABCDA$ . Part

$ABC$ is in verical plane

$PSTU$ while part

$CDA$ is in horizontal plane

$PQRS$ . Direction of circulation along the path is shown by an arrow near point

$B$ and

$D$ .

$\oint { \vec { B } .d\vec { l } }$ for this path according to Ampere's law will be :

0%
$\left( { I }_{ 1 }-{ I }_{ 2 }+{ I }_{ 3 } \right) { \mu }_{ 0 }$
0%
$\left( -{ I }_{ 1 }+{ I }_{ 2 } \right) { \mu }_{ 0 }$
0%
${ I }_{ 3 }{ \mu }_{ 0 }$
0%
$\left( { I }_{ 1 }+{ I }_{ 2 } \right) { \mu }_{ 0 }$

Explanation

$\mu_0(I_{enc})=\oint \bar B.\bar dl$

Area vector of ABC is along the right direction, while ACD is along the upward direction.

Now currents passing through ABC are

$I_1,I_3$ along the area directions.

Now through ACD,

$I_2,I_3; I_2$ along the area vector while

$I_3$ is opposite to it.

Thus,

$\mu_0(I_1+I_3+I_2-I_3) = \oint \bar B.\bar dl = \mu_0 (I_1+I_2)$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 12 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 12 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.