Exerts a force, if the charged particle is moving across the magnetic field line

Always exerts a force on charged particle

Never exerts a force on charged particle

Exerts a force, if the charged particle is moving along the magnetic field line

MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 10

The graph gives the magnitude B(t) of a uniform magnetic field that exist throughtout a conducting loop. perpendicular to the plane of the graph according to the magnitude of the emf induced in the loop greatest first

0%
$b > (d = \epsilon ) < (a = c )$ .
0%
$b > (d = \epsilon) > (a = c)$
0%
$b< d< \epsilon < \epsilon < c$
0%
$b> (a = c) > ( d = \epsilon )$

Explanation

Emf , $\varepsilon =\dfrac{AdB}{dt}$ .

$\varepsilon \,\,\alpha \,\,\dfrac{dB}{dt}\,$ (Slope of graph of magnetic field $B$ verses time $t$ )

From the diagram given in question.

In the given graph slope of $A B >$ slope of CD, slope in the $^ { \prime } a ^ { \prime }$ region slope in the $c ^ { \prime } c ^ { \prime }$ region $= 0 ,$ slope in the $' d ^ { \prime }$ region $=$ slope in the $^ { \prime } e ^ { \prime }$ region $\neq 0 .$

Hence, why $b > ( d = \varepsilon ) > ( a = c )$

${\vec B_1},{\vec B_2}$ and

${\vec B_3}$ are the magnetic field due to

$I_1 , I_2$ and

$I_3$ . Ampere's circuital law is given by

$\oint {\vec B} .d\vec l = {\mu _0}I$ , here

$\vec B$ is_____.

0%
$\vec B = {\vec B_1} - {\vec B_2}$
0%
$\vec B = {\vec B_1} + {\vec B_2} + {\vec B_3}$
0%
$\vec B = {\vec B_1} + {\vec B_2} - {\vec B_3}$
0%
$\vec B = {\vec B_1} - {\vec B_3}$

A long horizontal wire

$P$ carries a current

$50\ A$ . It is rigidly fixed. Another line wire

$Q$ is placed directly above and parallel to

$P$ . The weight of wire

$Q$ is

$0.075\ Nm^{-1}$ and caries a current of

$25\ A$ . Find the position of wire

$Q$ from

$P$ so that wire to remains suspended due to the magnetic repulsion-

0%
$3.33\ mm$
0%
$33.3\ mm$
0%
$334\ mm$
0%
$333\ mm$

Explanation

Formula,

$\dfrac{F}{l}=\dfrac{\mu _0}{4\pi} \dfrac{2I_1I_2}{r}$

$0.075=10^{-7}\times \dfrac{2\times 50\times 25}{r}$

$\Rightarrow r=3.33\times 10^{-3}$

$\therefore r=3.33mm$

A particle of mass

$1.6\times 10^{-27}\ kg$ and charge

$1.60\times 10^{-19}$ coulomb enters in a uniform magnetic field of

$1$ Tesla as shown in the fig. the speed of the particle is

$10^{7}\ m/s$ . The distance

$PQ$ will be-

0%
$0.14\ m$
0%
$0.28\ m$
0%
$0.4\ m$
0%
$0.5\ m$

Explanation

$\dfrac{m\times v}{qB}$

$=\dfrac{1-6\times 10^{-27}\times 10^7}{1.6\times 10^{-19}\times 1}$

$=10^{-1}$

$\Delta OPQ$

$PQ=2r\cos\theta$

$=\sqrt{2}\times 10^{-1}$

$=0.142m$

1280957_1081327_ans_8786905f817547a79869c29a60589ab6.png

As shown, a uniform magnetic filed B pointing out of the paper plane is confined in the shaded area of radius r. At a distance R from the center of the shaded area there is a point particle of mass m and carrying charge q. The magnetic filed is then quickly changed to zero.Find the speed of the particle.

0%
$v=\dfrac{qBr^2}{2mR}$
0%
$zero$
0%
$v=\dfrac{qBr^2}{mR}$
0%
$v=\dfrac{2qBr^2}{mR}$

A magnetic filed:

0%
Always exerts a force on charged particle
0%
Never exerts a force on charged particle
0%
Exerts a force, if the charged particle is moving across the magnetic field line
0%
Exerts a force, if the charged particle is moving along the magnetic field line

An electron has velocity v=

$\left( {2.0\times{{10}^6}\;m/s} \right)\hat{i} + \left( {3.0\times{{10}^{6\;}}m/s} \right)\hat{j}$ , magnetic field present is region is

$B\left( {0.030\;T} \right)\;i - \left( {0.15\;T} \right)j$ . Find the force on electron.

0%
$-6.24\times{10}^{-14} \hat{k}$
0%
$-0.624\times{10}^{-14} \hat{k}$
0%
$-62.4\times{10}^{-14} \hat{k}$
0%
$-624\times{10}^{-14} \hat{k}$

A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic vector field

$\overline {B}$ at a point having coordinates (x, y) in the z=0 plane is :

0%
$\dfrac{\mu_0I(y\hat{i} - x \hat{j})}{2\pi (x^2+y^2)}$
0%
$\dfrac{\mu_0I(x\hat{i} - y \hat{j})}{2\pi (x^2+y^2)}$
0%
$\dfrac{\mu_0I(x\hat{j} - y \hat{i})}{4\pi (x^2+y^2)}$
0%
$\dfrac{\mu_0I(x\hat{i} - y \hat{j})}{4\pi (x^2+y^2)}$

Explanation

An infinite wire carrying current in

$(-z)$ axis

we have

$\vec { l } =\left( -\hat { k } \right)$ (current vector)

$\vec { l } =x\hat { i } +y\hat { j } \Rightarrow { r }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }$

using Biot-Savart law fo rinfinite wire

$\vec { B } =\cfrac { { \mu }_{ 0 }I }{ 2\pi r } \left( \cfrac { \vec { l } \times \vec { r } }{ r } \right)$ (

$\cfrac { \vec { l } \times \vec { r } }{ r }$ is the unit vector along

$\vec { B }$ )

$=\cfrac { { \mu }_{ 0 }I }{ 2\pi { r }^{ 2 } } \left[ \left( -\hat { k } \right) \times \left( x\hat { i } +y\hat { j } \right) \right] \Rightarrow \vec { B } =\cfrac { { \mu }_{ 0 }I }{ 2\pi ({ x }^{ 2 }+{ y }^{ 2 }) } \left( y\hat { i } -x\hat { j } \right)$ (Ans)

1367638_1177298_ans_65408ca4a9824dcf97aa95664a5fb50b.PNG

A charged particle moving in a magnetic field experience maximum magnetic force when

0%
it moves parallel to the field
0%
it moves in opposite direction of the field
0%
it moves at right angles to the direction of field
0%
it moves at an angle $45^0$ to the direction of field

A charged particle moving in a magnetic field experiences, some magnetic force because

0%
Moving charged particle produce electric field
0%
Moving charged particle produce magnetic field
0%
Magnetic field do not allow the charged particle to pass through it.
0%
none of these

Motion of charges is noting but :

0%
Electric current
0%
magnetic effect
0%
heating effect
0%
all of the above

Explanation

Due to motion of charges through a conductor, the current of charges passing through its cross-section changes. It causes electric current

$\left( ie. \dfrac{da}{dt}\right)$ . Because of this current, heating effect and magnetic effect are produced in accordance with Joules and Bio Savaert's law resp.

Option

$-D$ is correct.

A particle enters a magnetic field perpendicular and undeviated, (there is no other field present ) then this partical must be

0%
$Electron$
0%
$Neutron$
0%
$\alpha-Particle$
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$Proton$

An electron is executing uniform circular motion in a uniform magnetic field

$\vec{B}$ . If magnetic moment is

$\vec{M}$ . Select incorrect option.

0%
$\vec{M}\times \vec{B}=0$
0%
$\vec{M}\cdot \vec{B} < 0$
0%
$\vec{M}\cdot \vec{B} > 0$
0%
Both $(1)$ & $(2)$

One gram hydrogen has

$6\times 10^{23}$ atoms. Imagine that all the nuclei are put at the north pole of the earth and the electrons at the south pole of the earth (radius=6400 km). The force between the charges is

0%
$10\times 10^{5}N$
0%
$5\times 10^{5}N$
0%
$2.5\times 10^{5}N$
0%
$2\times 10^{5}N$

Explanation

Given: Mass of hydrogen= 1 gram ;Radius=6400 km

To Find: Forces between charges

Solution : No of mole=

$\frac{weight}{molecular weight}=\frac{1}{2}$

No of electrons and protons=

$2*\frac{1}{2}*6*10^{23}$

$6*10^{23}$

Charge(q)=ne=

$1.6*10^{-19}*6*10^{23}$

=960000

Force=

$\frac{kq}{r^{2}}^{2}$

$\frac{9*10^{^{9}}*\left \{ \right.9.6*10^{4}\left. \right \}^{2}}{6400^{2}}$

$10*10^{5}$ N

Answer: Hence option A is correct

Two identical short bar magnets each of magnetic moment

$10\ Am^{2}$ are placed at a separation of

$10\ cm$ between their centres such that their axes are perpendicular to each other. The magnetic field at a point midway between the two magnets is

0%
$\sqrt{5} \times 10^{-6}\ T$
0%
$1.8\times 10^{-2}\ T$
0%
$4\times 10^{4}\ T$
0%
$12\times 10^{-4}\ T$

Explanation

Given: M=10A

$m^{2}$

Solution: When two identical short bar magnets are placed such that their axes are perpendicular to each other then that point will be axial position for first bar magnet and equatorial for another.

Formula used for magnetic field at axial position will be:

$B_{1}=\frac{\mu _{0} }{4\pi } \times \frac{2M }{r^{3}}$

Now, put values in the formula

$B_{1}=10^{-7} \times \frac{2\times 10 }{\left ( 0.05 \right )^{3}}$ =0.016

Formula used for magnetic field at equatorial position will be:

$B_{2}=\frac{\mu _{0} }{4\pi } \times \frac{M }{r^{3}}$

Now, put values in the formula

$B_{2}=10^{-7} \times \frac{ 10 }{\left ( 0.05 \right )^{3}}$ =0.008

Net magnetic field=

$\sqrt{B_{1}^{2}+B_{2}^{2}}$

$1.8\times 10^{-2}$

Three long, straight and parallel wires carrying currents are arranged as shown in the figure. The wire C Wh carries a current of 5.4 is so placed that it experience no force. The distance of wire C from wire D is the

0%
9 cm
0%
7 cm
0%
5 cm
0%
3 cm

Explanation

For no force on wire

$C$ ,

force on wire

$C$ due to wire

$D=$ force on wire

$C$ due to wire

$B$

Force on one wire due to the other current carrying wire is given by:

$F=\dfrac{\mu_0}{4\pi}\dfrac{2i_1i_2}{x}\times l$

So, it can be simplified as:

$\Rightarrow \dfrac{\mu _0}{4 \pi} \times \dfrac{2 \times 15 \times 5}{x} \times l=\dfrac{\mu _0}{4 \pi} \times \dfrac{2 \times 10 \times 5}{(15-x)} \times l$

$\Rightarrow x=9\ cm$

If the magnetic field at P can be written as K

$\left( \tan { \frac { a }{ 2 } } \right)$ then K is

0%
$\dfrac { { \mu }_{ 0 }I }{ 4\pi d }$
0%
$\dfrac { { \mu }_{ 0 }I }{ 2\pi d }$
0%
$\dfrac { { \mu }_{ 0 }I }{ \pi d }$
0%
$\dfrac { {2 \mu }_{ 0 }I }{ \pi d }$

Fleming's left hand rule is used to find

0%
Direction of magnetic field
0%
Direction of current
0%
Direction of magnetic force acting on conductor
0%
None of these

Three plotting compasses are placed close to a current carrying wire wrapped around an insulator as shown in the figure.
How many compass needles will change direction if the current through the wire is increased? (Ignore the effect of the earth's magnetic field.)

0%
0
0%
1
0%
2
0%
3

The magnetic moment of a current carrying loop is

$2.1\times {10}^{-25}$

$amp\times {m}^{2}$ . The magnetic field at a point on its axis at a distance of

$1\ A^o$ is

0%
${ 4.2\times 10 }^{ -2 } \ weber/{m}^{2}$
0%
${ 4.2\times 10 }^{ -3 }\ weber/\ {m}^{2}$
0%
${ 4.2\times 10 }^{ -4 }\ weber/\ {m}^{2}$
0%
${ 4.2\times 10 }^{ -5 }\ weber/\ {m}^{2}$

Explanation

Field at a point

$x$ from the centre of a current carrying loop on the axis is

$B=\dfrac{\mu_0}{4\pi}.\dfrac{2M}{x^3}$

$=\dfrac{10^{-7}\times 2\times 2.1 \times 10^{-25}}{(10^{-10})^3}$

$=4.2 \times 10^{-32}\times 10^{30}$

$=4.2 \times 10^{-2}\ W/m^2$

A current is flowing through a thin cylindrical shell of radius R. If the energy density in the medium, due to the magnetic field, at a distance r from the axis of the shell is equal to U then which of the following graphs is correct?

Explanation

When a current flows through cylindrical shell, then according to Ampere circuital law, magnetic induction inside it will be equal to zero. Hence energy density at

$r<R$ is eual to zero.

Therefore,

$(a),(c)$ and

$(d)$ are wrong.

When

$r>R, B=\dfrac{\mu_{0}i}{2\pi r}$

since

$U=\dfrac{B^{2}}{2\mu_{0}},$ therefore, outside the shell,

$U=\dfrac{\mu_{0}i^{2}}{9\pi^{2}r^{2}}.$

It means, just outside the shell,

$U=\dfrac{\mu_{0}i^{2}}{9\pi^{2}R^{2}}$ and when

$r\rightarrow \infty, U\rightarrow 0.$

Hence

$(b)$ is correct.

A conductor of length

$1\ m$ and carrying current of

$1\ A$ is placed at an angle

$45 ^ { \circ }$ to the magnetic field of 1 oersted. The force acting on conductor is

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

Explanation

It is given that,

Length, $l=1\,m$

Current, $I=1\,A$

Magnetic field, $B=1\,oersted={{10}^{-4}}\,Kg/A{{s}^{2}}$

Magnetic force,

$F=I\left( L\times B \right)$

$F=1\left( 1\times {{10}^{-4}}\sin 45 \right)$

$F=\dfrac{{{10}^{-4}}}{\sqrt{2}}$

Two magnetic bars are suspended in horizontal plane and ocillates in earth magnetic field. Their time period of oscillations are

$T_1$ and

$T_2$ . The ratio of their magnetic dipole moments is :

0%
$\dfrac{T_1}{T_2}$
0%
$\dfrac{T_2}{T_1}$
0%
$\left(\dfrac{T_1}{T_2}\right)^2$
0%
$\left(\dfrac{T_2}{T_1}\right)^2$

Explanation

Given :

Time period of two magnetic bars as

$T_1$ and

$T_2$ .

Solution:

From angular SHM and torque relation

$\tau = \overrightarrow{M} \times \overrightarrow{B}$

Time period of oscillation is given as :

$T = 2\pi \sqrt{\dfrac{I}{MB}}$

So,

$T \; \alpha \; \sqrt{\dfrac{1}{M}}$

So,

$\dfrac{T_1}{T_2} = \sqrt{\dfrac{M_2}{M_1}}$

Square both sides,

Hence,

$\dfrac{M_1}{M_2} = \left(\dfrac{T_2}{T_1}\right)^{2}$

Correct Opt:[D]

ON connecting a battery to the two opposite corner of a diagonal of a square wire frame of side 10 cm.The magnetic field at the centre will be (wire is uniform)

0%
zero
0%
1 $\mu{T}$
0%
2 $\mu{T}$
0%
4 $\mu{T}$

Explanation

$\textbf{Solution}:$

Magnetic field due to different side,

Due to AB,

$B_1=\dfrac{\mu_o}{4\pi}\dfrac{i}{10}[sin45^o+sin45^o](inward)$

Due to BC,

$B_2=\dfrac{\mu_o}{4\pi}\dfrac{i}{10}[sin45^o+sin45^o](inward)$

Due to CD,

$B_3=\dfrac{\mu_o}{4\pi}\dfrac{i}{10}[sin45^o+sin45^o](outward)$

Due to DA,

$B_4=\dfrac{\mu_o}{4\pi}\dfrac{i}{10}[sin45^o+sin45^o](outward)$

Now,

$B_{net}=B_1+B_2+B_3+B_4$ = 0

$\textbf{Hence A is the correct option}$

Two concentric circular coil of radius

$20 cm$ and

$30 cm$ carries current

$2A$ and

$3A$ respectively in opposite direction then magnetic field at centre will be :-

0%
$4\pi \times 10^{-7}$
0%
$2\pi \times 10^{-7}$
0%
$2 \times 10^{-7}$
0%
$0$

Explanation

$0$

Given,

Two concentric circular coils of radius

$r_1= 20cm$ carry current

$I_1=2A$

$r_2= 30cm$ carry current

$I_2=3A$

$\mu_0=4\pi\times 10^{-7} TmA^{-1}$

The magnetic field at the Centre of a circular coil of radius

$r$ is,

$B=\dfrac{\mu_0I}{2r}$

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

$r_1$ ,

$B_1=\dfrac{\mu_0 I_1}{2r_1}$

$B_1=\dfrac{4\pi\times10^{-7}\times2}{2\times20\times 10^{-2}}$

$=2\pi\times10^{-6}T$

The magnetic field at the Centre of a coil of radius

$r_2$ ,

$B_2=\dfrac{\mu_0I_2}{2r_2}$

$B_2=\dfrac{4\pi\times10^{-7}\times3}{2\times30\times10^{-2}}$

$= 2\pi\times10^{-6}T$

Both current

$I_1$ and

$I_2$ flow in the opposite direction, Direction of a magnetic field is also in the opposite direction.

So, Total magnetic field at the Centre of two concentric coils is

$B_1-B_2= 0T$

A wire is bent into three successive quadrants. The quadrant

$ab$ lies in the

$xy$ plane.

$b$ in

$xy$ plane and

$ca$ in the

$zx$ plane. What is the magnetic moment of this system if a current

$I$ flows through it? Give:

$R=$ radius each quadrant.

0%
$\dfrac {\sqrt {3}\pi r^{2}I}{4}$
0%
$\dfrac {\sqrt {2}\pi r^{2}I}{4}$
0%
$\dfrac {\pi r^{2}I}{2}$
0%
$\dfrac {\pi r^{2}I}{4}$

Explanation

Given: current=I ,radius=R

Solution: Let us suppose that the wire in each quadrant forms a loop whose

magnetic moment we have to find :

$\vec{M}=I\times \frac{\pi r^{2}}{4}(\hat{k})+I\times \frac{\pi r^{2}}{4}\hat{(j)}+I\times \frac{\pi r^{2}}{4}(\hat{i})$

Remember direction is given by right hand thumb rule.

$|\vec{M}|=\sqrt{(I\times \frac{\pi r^{2}}{4})^{2}+(I\times \frac{\pi r^{2}}{4})^{2}+(I\times \frac{\pi r^{2}}{4})^{2}}$

$M=\sqrt{3}\times I\times \frac{\pi r^{2}}{4}$

Hence the correct option is A

A charged particle enters in a magnetic field

$H$ with its initial velocity making an angle of

$45^o$ with

$H$ . The path of the particle will be

0%
an ellipse
0%
straight line
0%
a circle
0%
a helical

Explanation

The component of velocity perpendicular to

$H$ will make the motion circular while that parallel to

$H$ will make it move along a straight line. The two together will make the motion helical.

In fig. two long parallel wires carry equal currents in opposite directions. Point O is situated mid way between the wires and X - Y plane contains two wires and +ve Z - axis comes normally out of the plane of paper. The magnetic field B at O is non - zero along

0%
X Y and Z axes
0%
X - axis
0%
Y - axis
0%
Z - axis

Explanation

$Solution :$ From Right hand rule, magnetic field at

$O$ due to wire on the left is inside the paper and due to the wire on the right is also inside the paper.

Hence, no component of magnetic field is in

$x$ or

$y$ direction and the only remaining component is in

$z$ -direction.

$So, the$

$correct$

$option: D$

If the angular momentum of an electron is

$\overset{\rightarrow}{J}$ then the magnitude of the magnetic moment will be

0%
$\cfrac{eJ}{m}$
0%
$\cfrac{2m}{eJ}$
0%
$eJ2m$
0%
$\cfrac{eJ}{2m}$

Explanation

$Given :$ Angular momentum of an electron =

$\vec {J}$

$Solution :$ The magnetic momment of an electgron revolving in a circular orbit of radius

$r$ is

$M$ =

$1.2evr$

And its angular momentum

$J$ =

$Iw$ =

$mr^2$

$\times$

$\dfrac {v} {r}$ =

$mvr$

$\therefore$

$\dfrac {M} {J}$ =

$\dfrac {1} {2}$

$\dfrac{evr} {mvr}$ =

$\dfrac {1} {2}$

$\dfrac {e} {m}$

$\therefore$

$M$ =

$\dfrac {eJ} {2m}$

$So,the$

$correct$

$option : D$

A conducting rod

$MN$ of mass

$'m$ and length

$'\ell '$ is placed on parallel smooth conducting rails connected to an uncharged capacitor of capacitance

$'C'$ and a battery of emf

$\varepsilon$ as shown. A uniform magnetic field

$'B'$ is existing perpendicular to the plane of the rails.The steady state velocity acquired by the conducting rod

$MN$ after closing switch

$S$ is (neglect the resistance of the parallel rails and the conducting rod )

0%
$\frac{{2CB\ell \varepsilon }}{{\left( {m + C{B^2}{\ell ^2}} \right)}}$
0%
$\frac{{CB\ell \varepsilon }}{{\left( {m + C{B^2}{\ell ^2}} \right)}}$
0%
$\frac{{CB\ell \varepsilon }}{{2\left( {m + C{B^2}{\ell ^2}} \right)}}$
0%
$\frac{{CB\ell \varepsilon }}{{4\left( {m + C{B^2}{\ell ^2}} \right)}}$

Explanation

Given :Mass=m ,Length =l, Capacitance=C, Emf=

$\varepsilon$ ,Magnetic field =B

Solution: Let the potential difference across conducting rod=

$V_{1}$

Let the potential difference across capacitor=

$V_{2}$

So,

$\varepsilon=V_{1}+V_{2}$

$\varepsilon=Bvl+\frac{q}{C}$

$q=\left ( \varepsilon -Bvl \right )C$ ......(1)

Force required to pull the loop with constant velocity:

$F=ma=m\frac{dv}{dt}$

$Bil=m\frac{dv}{dt}$

Integrating the euation:

$m\int_{0}^{v}dv=Bl\int_{0}^{q}dq\times dt$

$m\left ( v-0 \right )=Blq$

$q=\frac{mv}{Bl}$ .....(2)

From equation (1) & (2):

$\left ( \varepsilon -Bvl \right )C=\frac{mv}{Bl}$

The steady state velocity acquired by the conducting rod MN after closing switch S is:

$v=\frac{C\varepsilon Bl}{m+B^{2}l^{2}C^{2}}$

Hence the correct option is B

Two point dipoles of dipole moment

$\xrightarrow [ P_ 1 ]{ } \quad and\quad \xrightarrow [ P_ 2 ]{ }$ are at a distance x from each other and

$\xrightarrow [ P_ 1 ]{ } \parallel \xrightarrow [ P_ 2 ]{ }$ The force between the dipoles is:

0%
$\dfrac { 1 }{ 4\pi e_ 0 } \dfrac { 4p_ 1p_ 2 }{ x^ 4 }$
0%
$\dfrac { 1 }{ 4\pi e_ 0 } \dfrac { 3p_ 1p_ 2 }{ x^ 3 }$
0%
$\dfrac { 1 }{ 4\pi e_ 0 } \dfrac { 6p_ 1p_ 2 }{ x^ 4 }$
0%
$\dfrac { 1 }{ 4\pi e_ 0 } \dfrac { 8p_ 1p_ 2 }{ x^ 4 }$

Explanation

Given

Two parallel dipoles at distance x

Solution

Axial electric field of dipole P1

$E_{P_{1}}=\frac{2*P_{1}}{4\pi \epsilon _{0}*x^{3}}$

Force on dipole P2 due to electric field of P1

$F=P_{2}*\frac{dE}{dx}$

$P_{2}*\frac{d}{dx}(\frac{2P_{1}}{4\pi \epsilon _{0}*x^{3}})$

On solving

$F=-(\frac{6P_{1}P_{2}}{4\pi \epsilon _{0}*x^{4}})$

The correct option is C

1998650_1337139_ans_82f3fbe8ed8d426abbed94dac8b87440.jpeg

A magnetised wire of moment

$3.14\ A-m^{2}$ is bent in the form of a semi-circle ; then the new magnetic moment will be:-

0%
$3.14\ A-m^{2}$
0%
$2\ A-m^{2}$
0%
$6.28\ A-m^{2}$
0%
$Noone\ of\ these$

Explanation

The pole strength of wire is assumed to be

$P$ and the length is assumed

$L$ .

Then,

$\Rightarrow$

$P\times L=3.14\,Am^2$

The wire is then bent in the form of a semicircle.

We now,

$L=3.14R$

$\Rightarrow$ The semi-circle effective length

$=2R=\dfrac{2L}{3.14}$

$\Rightarrow$ The new magnetic moment formed in wire

$=P\times (2R)$

$=\left(\dfrac{2L}{3.14}\right)\times P$

$=\dfrac{2}{3.14}\times 3.14$ [ Since,

$P\times L=3.14$ ]

$=2\,A-m^2$

Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper, as shown in figure. The variation of the magnetic field B along the line XX' is given by :-

The intensity of magnetic field is H and moment of magnet is M. The maximum potential energy is

0%
MH
0%
2 MH
0%
3 MH
0%
4 MH

Explanation

When the magnetic field

$=H$

and moment of magnet

$=M$

Now, the maximum potential energy

$=HM$

when

$\sin{ 90 }^{ 0 }=1$

Density of iron is 7800 kg/m

$^{3}$ and induced magnetic field in iron is 8 x10

$^{5}$ A/m then find magnetic dipole moment of each iron atom is (nearly)

0%
$6\times10$ ^{-24} $Am$ ^{2} $$
0%
$9\times10$ ^{-22} $Am$ ^{2} $$
0%
10 $^{-23}$ Am $^{2}$
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$4\times10$ ^{-24} $Am$ ^{2} $$

Explanation

Given: density of iron,

$\rho=7800 kg/m^3$

induced magnetic field,

$M=8*10^5A/m$

To find: dipole moment,

$m=?$

Solution: As we know that,

$mass of iron=56a.m.u.=56*1.66*10^{-27}kg$

So, volume of iron,

$V=\dfrac{mass of iron}{\rho}$

$==>V=\dfrac{56*1.66*10^{-27}}{7800}$

$==>V=0.01191794872*10^{-27}m^3$

As we know that,

$magnetisation=\dfrac{magnetic moment}{volume}$

$==>M=\dfrac{m}{V}$

$==>m=MV$

$==>m=8*10^5*0.01191794872*10^{-27}$

$==>m=0.0953*10^{-22}$

$==>m\simeq 0.1*10^{-22}Am^2$

$==>m\simeq 10^{-23}Am^2$

hence,

The correct opt: C

If two streams of electrons move parallel to each other in same direction then they.

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attract each other
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repel each other
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do not exert any force on one other
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Get rotated perpendicular to each other

An infinitely straight conductor

$AB$ is fixed and a current is passed through it Another movable straight wire

$CD$ of finite length and carrying current is held perpendicular to it and released. Neglect weight of the wire

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The rod $CD$ will move upwards parallel to itself
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The rod $CD$ will move downward parallel to itself
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The rod $CD$ will move upward and turn clockwise at the same time
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The rod $CD$ will move upward and turn anticlockwise at the same time

Explanation

Since the force on the rod

$CD$ is non-uniform it will experience force and torque. From the left hand side it can be seen that the force will be upward and torque is clockwise.

Correct option

$(c)$ . The rod CD will move upward and turn clockwise at the same time

1727486_1456866_ans_f9eb05408d194f27978ec04784ebf8e5.png

A thin circular disk of radius R is uniformly charged with density

$\sigma>0$ per unit area. The disk rotates about axis with a uniform angular speed

$\omega$ . The magnetic moment of the disk is:

0%
$2 \pi R^{4} \sigma \omega$
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$\pi R^{4} \sigma \omega$
0%
$\dfrac{\pi R^{4}}{2} \sigma w$
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$\dfrac{\pi R^{4}}{4} \sigma \omega$

Explanation

Let us consider the disc to be made up of a larger number of concentric elementary rings.

Consider one such ring of radius

$x$ and thickness

$dx$

Charge on this elementary ring,

$dq=\sigma\times 2\pi xdx=2\pi \sigma xdx$

Current associated with this elementary ring,

$dl=\dfrac{dq}{dt}=dq\times f=\sigma \omega x dx$

Here

$f$ is the frequency and

$\omega=2\pi f$

Magnetic moment of this elementary ring,

$dM=dl\pi x^2=\pi\sigma \omega x^3 dx$

Therefore, magnetic moment of the entire disc,

$M=\int_0^R=\pi \sigma \omega \int_0^R x^3 dx=\dfrac{\pi R^4}{4}\sigma \omega$

1475584_1439284_ans_5c4f9a8d6c4642f498c22f5392642e53.png

Two particles having same charge and KE enter at right angles into the same magnetic field and travel in circular paths of radil

$2 \mathrm{cm}$ and

$3 \mathrm{cm}$ respectively. The ratio of their masses is.

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$2 : 3$
0%
$3 : 2$
0%
$4 : 9$
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$9 : 4$

Explanation

$\textbf{Given}:$

$r_{1}=2cm$ ,

$r_{2}=3cm$

$\textbf{Solution}:$
We know,

$r= \dfrac {\sqrt {2mqV}}{qB}$
Hence,

$\sqrt {\dfrac {2mV}{aB^{2}}}$

$\dfrac {r_{1}}{r_{2}}= \sqrt {\dfrac {m_{1}}{m_{2}}}$

$\dfrac {m_{1}}{m_{2}}= \dfrac {r_{1}^{2}}{r_{2}^{2}}$
Hence,

$\dfrac {m_{1}}{m_{2}}= \dfrac {(2)^2}{(3)^2}= \dfrac {4}{9}$

$\textbf{Hence C is the correct option}$

A fixed horizontal wire carries a current of

$200 A.$ Another wire having a mass per unit length

$10^{3} kg/m$ is placed below the first wire at a distance of 2 cm and parallel to it How much current must be passed through the second wire if it floats in air without any support? What should be the direction of current in it

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25A (direction of current is same to first wire)
0%
25A (direction of current is opposite to first wire)
0%
49A (direction of current is same to first wire)
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49A (direction of current is opposite to first wire)

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

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

$\vec{a}=2\hat{i}+x\hat{j}$ . Select the correct alternative(s) :

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$x = 1.5$
0%
$x = 3$
0%
magnetic field is along z-direction
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kinetic energy of the particle is constant

Explanation

Force & velocity are perpendicular to each other in magnetic field
So

$\vec{V}\cdot\vec{a}=0$

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

$6+4x=0$

$x=-1.5$
No work is done by magnetic field. So, KE of particle is constant

A charge particle of mass m and charge q is projected with velocity v along plane at a distance a from a long staright current carrying conductor as shown in figure. the radius of curvature of the path traced by the particle at the given position does not depend on

0%
$\theta$
0%
q
0%
i
0%
m

Explanation

$Given:$ mass=

$m$ ; charge=

$q$ ; velocity =

$v$ ; distance=

$a$

$Solution:$ radius of curvature

$r=\dfrac{mV} {qB}$

So,

$B = \dfrac {\mu _o i} {2 \pi a}$

$r = \dfrac{mV 2 \pi a} {q \mu _ o i}$

so, independent of

$\theta$

$So,the$

$correct$

$option:A$

A coil in the shape of an equilateral triangle of side 0.02 m is suspended from its vertex such that it is hanging in a vertical plane between the poles pieces of permanent magnet producing a uniform field of

$5 \times 10^{-2}$ T. If a current of 0.1 A is passed through the coil, what is the couple acting

0%
$5\sqrt{3} \times 10^{-7}$ N-m
0%
$5\sqrt{3} \times 10^{-10}$ N-m
0%
$\dfrac{\sqrt{3}}{5} \times 10^{-7}$ N-m
0%
None of these

Explanation

$Given:$ Side

$=0.02m$ ;

$I=0.1A$ ;

$B=5\times10^{-2} T$

$Solution:$ As the coil is in the form of an equilateral triangle, its

area

$\begin{aligned} A &=\frac{\sqrt{3}}{4} \times 0.02 \times 0.02 \\ A &=\sqrt{3} \times 10^{-4} m \\ &=\sqrt{3} \times 10^{-4} m ^{2} \end{aligned}$

Torque on current carring coil in magnetic field is

$T=$

$BINA sin\alpha$

$\alpha=90^{\circ}$

$N =1$

$T = BIA$

$=5 \times 10^{-2} \times 0.1 \times \sqrt{3} \times 10^{-4} N ^{- m }$

$=0.5 \times \sqrt{3} \times 10^{-6}$

$=5 \sqrt{3} \times 10^{-7}$

$N^{-m}$

$So,the$

$correct$

$option:A$

2004498_1553286_ans_0498951f6974467daa303a9c0c8a864b.jpeg

An iron rod of length L and magnetic moment M is bent in form of a semicircular.Now its magnetic moment will be.

0%
M
0%
$\frac {2M}{\pi}$
0%
$\frac {M}{\pi}$
0%
$M \pi$

Explanation

Given: Length of iron rod = L

Magnetic moment of rod = M

Solution: As we know,

Magnetic moment is given by,

$M=m\times L$ (1)

According to question, when we bent the rod the pole strength of the rod remains unchanged. However, when the rod is bent in form of a semicircular arc the separation between the two poles become 2r (r is the radius of the semicircular arc).

$\therefore \pi r=L$

$r=\frac{L}{\pi }$

Therefore, the new magnetic moment will be,

$M'=m\times 2r$

$M'=m\times \frac{2L}{\pi}$

$=\frac{2M}{\pi }$

Hence, the correct option is (B).

The Biot-Savart's law in vector from is:

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$d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { di\left( \overrightarrow { l } \times \overrightarrow { r } \right) }{ r^ 2 }$
0%
$d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { dl } \times \overrightarrow { r } \right) }{ r^ 2 }$
0%
$d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { r } \times \overrightarrow { dl } \right) }{ r^ 2 }$
0%
$d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { dl } \times \overrightarrow { r } \right) }{ r^ 3 }$

Explanation

Solution: The law states that the magnetic field induction at any point due to a current

$I$ in a small element of length

$dl$ at a distance

$r$ from the element is given by,

$dB=\frac{\mu _{0}Idl \sin \theta }{4\pi r^{2}}$

$\overrightarrow{dB}=\frac{\mu _{0}I(\overrightarrow{dl}\times \overrightarrow{r})}{4\pi r^{3}}$

Hence, the correct option is (D).

A circular loop of area

$1\ cm^2$ , carrying a currrent of

$10$ A, is placed in a magnetic field of

$0.1$ T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is

0%
zero
0%
$10^{-4}\ \hbox{N-m}$
0%
$10^{-2}\ \hbox{N-m}$
0%
$1\ \hbox{N-m}$

Explanation

Given: Area of the loop=

$1 cm^{2}$

Current in the loop=

$10A$

Magnetic field=

$0.1T$

Solution: The torque on the loop due to the magnetic field is given by,

$\tau =NBiA\sin \theta$

where, B is the magnetic field,

A is the area of loop,

i denotes the current in the loop.

As,

$\theta = 0^{\circ}$

$\sin 0^{\circ}=0$

So,

$\tau$ =0

Hence, the correct option is (A).

Two similar bar magnets P and Q, each of magnetic moment

$\mathrm{M}$ , are taken. If P is cut along its axial line and Q is cut along its equatorial line, All the four pieces obtained have

0%
equal pole strength
0%
magnetic moment $\dfrac M4$
0%
magnetic moment $\dfrac M2$
0%
magnetic moment $M$

Explanation

If pole strength, magnetic moment and length of each part are

$m' ,M'$ and

$L'$ respectively then

$L'=\dfrac{L}{2}$

$\Rightarrow M' =\dfrac{M}{2}$

1737876_1506745_ans_416595fd497a4765bc74458a690ddcee.png

Two infinitely long parallel wires carry equal current in same direction. The magnetic field at a mid point in between the two wires is

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Twice the magnetic field produced due to each of the wires
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Half of the magnetic field produced due to each of the wires
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Square of the magnetic field produced due to each of the wires
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Zero

Explanation

At midpoint, magnetic fields due to both the wires are equal and opposite. So

$B=0$ .

A charge

$'q'$ moves in a region where electric field

$'E'$ nd magnetic filed

$'B'$ both exist, then force on it is :-

0%
$q(\vec{V}\times \vec{B})$
0%
$q\vec{E}$
0%
$q\left\{\vec{E}+(\vec{v}\times \vec{B}) \right\}$
0%
$q\left\{\vec{B}+(\vec{v}\times \vec{E}) \right\}$

Explanation

Solution: While moving in a space, the force experienced by the charged particle in the region where magnetic field and electric field exists is called Lorentz Force.

The force experienced by a charge particle of charge "q" moving in an electric field is given by,

$\overrightarrow{F_{e}}=q\overrightarrow{E}$

The direction of force depends on the nature of charge. If positive, the direction of force is in the direction of electric field, and if negative the direction of force is opposite to the direction of electric field.
The force experienced by a charge particle moving in the magnetic field

$\overrightarrow{B}$ with velocity

$\overrightarrow{v}$ is given by,

$\overrightarrow{F_{m}}=q(\overrightarrow{v}\times \overrightarrow{B})$

The direction of force is in the direction of

$(\overrightarrow{v}\times \overrightarrow{B})$ given by right hand screw rule.

So, the total force experienced by charge particle is given by,

$\overrightarrow{F}= \overrightarrow{F_{e}}+\overrightarrow{F_{m}}=q(\overrightarrow{E}+(\overrightarrow{v}\times \overrightarrow{B}))$

Hence, the correct option is (C).

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. let

$r_p$ ,

$r_e$ and

$r_{He}$ be their respectively radii, then ,

0%
$r_e > r_p > r_{He}$
0%
$r_e < r_p < r_{He}$
0%
$r_e < r_p = r_{He}$
0%
$r_e > r_p = r_{He}$

Explanation

$r=\dfrac{mv}{qB}=\dfrac{\sqrt{2mK}}{qB}$

$r_{He}=r_p > r_e$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 10 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 10 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

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