MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 8

If a current carrying wire carries 10A current then the magnetic field is X. Now the current in the wire increases to 100A, them magnetic field in the wire becomes

0%
>X
0%
<X
0%
=X
0%
all

What is shape of magnet in moving coil galvanometer to make the radial magnetic field?

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Concave
0%
Horse shoe magnet
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Convex
0%
None of the above

The force a magnetic field exerts on an electron is largest when the path of the electron is oriented

0%
In the opposite direction from the magnetic field's direction
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In the same direction as the magnetic field's direction
0%
Up through the magnetic field at a $45^{\circ}$ angles
0%
Down through the magnetic field at $45^{\circ}$ angle
0%
At a right angle to the magnetic field

Explanation

The magnetic force on an electron

$e$ , moving with velocity

$v$ in a magnetic field

$B$ is given by ,

$F=evB\sin\theta$ , where

$\theta$ is the angle between

$v$ and

$B$ ,

force

$F$ will be maximum when

$\sin\theta$ is maximum ,

and maximum value of

$\sin\theta$ is

$1$

i.e.

$\sin\theta=1=\sin90$

$\theta=90$

it means that magnetic force will be maximum when electron is moving at right angle to the magnetic field.

Positively charged particles are projected into a magnetic field. If the direction of the magnetic field is along the direction of motion of the charge particle, the particles get

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Accelerated
0%
Decelerated
0%
Delfected
0%
No changed in velocity

Explanation

The direction of magnetic field is along the direction of motion of the charge particles, so angle will be

$0^{\circ}$ .

$\therefore$ Force

$F = qvB\sin \theta$

$= qvB\sin 0$

$= 0 (\because \sin 0 = 0)$
So, there will be no change in the velocity.

Which of the following relation represents Biot-Savart's law?

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$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ r }$
0%
$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \hat { r } }{ { r }^{ 3 } }$
0%
$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 3 } }$
0%
$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 4 } }$

Explanation

If unit current is flowing through the conductor, then Biot-Savart's law is represented as

$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 3 } }$

A particle with velocity

$2\times10^4m/s$ enters the uniform magnetic field perpendicularly. Calculate the magnitude of the magnetic force on this particle if charge of particle is -0.04 C and magnetic field of strength is B= 0.5 T.

0%
4 N
0%
8 N
0%
40 N
0%
80 N
0%
400 N

Explanation

Magnetic force

$F = q (v \times B)$

$v \perp B$

$\implies |F| =|q|vB$

$\therefore$

$|F| = 0.04 \times (2\times 10^4) \times 0.5 = 400$

$N$

Which of the following material is used in making the core of a moving coil galvanometer?

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Copper
0%
Nickel
0%
Iron
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Both (a) and (b).

Three long straight wires A, B and C are carrying currents as shown in figure. Then the resultant force on B is directed :

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perpendicular to the plane of the paper and outward
0%
perpendicular to the plane of the paper and inward
0%
towards A
0%
towards C

Explanation

Hint:

If the current in the two parallel wires is in the same direction then the wires attract each other.

Step 1: Expression of force between two parallel wires,

The force between two parallel wires having current

$I_1$ and

$I_2$ and separated by distance

$d$ is,

$F=\frac{\mu_oI_1 I_2}{2\pi d}$

Therefore,

$F\propto I_1I_2$ and

$F\propto \frac{1}{d}$

Step 2: Find the stronger force.

Wire A and wire C are trying to pull wire B towards them.

The separation between the wires for both cases is the same.

$I_A I_B=(1A)(2A)=3A^2$

$I_c I_B=(3A)(2A)=6A^2$

$\because$

$F\propto I_1I_2$

$\therefore$

$F_{BC}>F_{AB}$

Therefore wire is pulled towards

$C$ .

$\textbf{Hence option D correct}$

A pair of long, straight current-carrying wires and four marked points are shown in above figure. Find out the points at which net magnetic field is zero?

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Point 1 only
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Points 1 and 2 only
0%
Point 2 only
0%
Points 3 and 4 only
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Point 3 only

Explanation

Magnetic field at a point at a distance

$d$ from a current carrying wire is

$\dfrac{\mu_0I}{2\pi d}$

Thus

$B\propto \dfrac{I}{d}$

At point 1,

Magnetic field due to wire with current

$I=\dfrac{\mu_0I}{2\pi (3a)}$ out of the plane

Magnetic field due to wire with current

$2I=\dfrac{\mu_0 (2I)}{2\pi (6a)}=\dfrac{\mu_0I}{2\pi(3a)}$ into the plane

Thus net magnetic field at point 1 is zero.

A magnet of magnetic moment 20 CGS units is freely suspended in a uniform magnetic field of intensity 0.3 CGS units. The amount of work done in deflecting it by an angle of

$30^o$ in CGS units is

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6
0%
$3\,\overline{3}$
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$3(2-\overline{3})$
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3

Explanation

Work done,

$W =-MB_{\mu} (1 - cos\theta )$

$= 20 \times 0.3(1 - cos 30^o)$

$=6\,1-\frac{\overline{3}}{2}=3(2-\overline{3})$

Match the following and find the correct pairs:

List I	List II
(a) Fleming's left hand rule	(e) Direction of induced current
(b) Right hand thumb rule	(f) Magnitude and direction of magnetic induction
(c) Biot-Savart law	(g) Direction of force due to magnetic induction
(d) Fleming's right hand rule	(h) Direction of magnetic lines due to current

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a-g, b-e, c-f, d-h
0%
a-g, b-h, c-f, d-e
0%
a-f, b-h, c-g, d-e
0%
a-h, b-g, c-e, d-f

Two identical concentric coils

$X$ and

$Y$ carrying currents in the ratio

$1:2$ are arranged in mutually perpendicular planes. If the magnetic field due to coil

$X$ is

$B$ the net field at their common centre is:

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$B$
0%
$2B$
0%
$3B$
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$\sqrt{5} B$

Explanation

As the coil are arranged in mutually perpendicular planes, the field produced at their common centre

${B}_{X}$ and

${B}_{Y}$ will be perpendicular to each other.

More over, except current, all other parameters being same.

$\dfrac { { B }_{ X } }{ { B }_{ Y } } =\dfrac { { I }_{ X } }{ { I }_{ Y } } =\dfrac { 1 }{ 2 }$

$\Rightarrow { B }_{ X }=B$ &

${ B }_{ Y }=2B$
The net magnetic field is given by,

${ B }_{ net }=\sqrt { { B }_{ X }^{ 2 }+{ B }_{ Y }^{ 2 } } =\sqrt { { B }^{ 2 }+{ \left( 2B \right) }^{ 2 } } =\sqrt { 5 } B$

A point charged particle of mass

$2\times 10^{-4} kg$ is moving perpendicular to the uniform magnetic field of magnitude 0.1-tesla.Calculate the acceleration of the particle due to the magnetic field if its velocity is

$3\times 10^4m/s$ and its charge is

$+4.0\times 10^{-9}C$ .

0%
0.0006 $m/s^2$
0%
0.006 $m/s^2$
0%
0.06 $m/s^2$
0%
0.6 $m/s^2$
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None of the above

Explanation

If a be the acceleration of the particle due to only magnetic field B , then

$ma=qvB$

$a=\dfrac{qvB}{m}$

$=\dfrac{(4\times 10^{-9})\times (3\times 10^4)\times 0.1}{2\times 10^{-4}}$

$=0.06 m/s^2$

As shown above, a magnetic force of

$10^{-14}$ N is experienced by a charge particle moving with speed of

$10^6$ m/s in a uniform magnetic field B of magnitude

$10^{-2}$ T. Calculate the magnitude of the charge.

0%
$10^{-22}$ C
0%
$10^{-18}$ C
0%
$10^{-10}$ C
0%
$10^{-6}$ C
0%
$10^{-2}$ C

Explanation

Given :

$F_m =10^{-14}$ N ;

$B = 10^{-2}$ T ;

$v = 10^6$ m/s

Using

$F_m = |q|vB$

$\therefore$

$10^{-14} = |q| \times 10^6 \times 10^{-2}$

$\implies$

$|q| =10^{-18}$ C

A charged particle is moving in circular path of radius 100 meters in a uniform magnetic field. If mass of the charged particle is

$9.1$ x

$10^{-31}$ kg and its speed is 3 x

$10^7$ m/s. Which of the following is the best estimate of the order of magnitude of the magnetic force needed to maintain this orbit?

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$10^{-22}$
0%
$10^{-17}$
0%
$10^{-10}$
0%
$10^{-6}$
0%
$10^{-2}$

Explanation

Given :

$r = 100$ m;

$m = 9.1 \times 10^{-31}$ kg;

$v = 3 \times 10^7$ m/s

Magnetic force

$F = \dfrac{mv^2}{r}$

$= \dfrac{(9.1 \times 10^{-31}) \times (3\times 10^7)^2}{100}$

$=8.2 \times 10^{-18} \approx 10^{-17}$ N

A proton enters a region of constant magnetic field,

$B$ , of magnitude

$1.0$ tesla with initial speed of

$1.5 \times {10}^{6} {m}/{s}$ . If the protons initial velocity vector makes an angle of

${30}^{o}$ with the direction of

$B$ . Calculate the speed of the protons

$4$ seconds after entering the magnetic field.

0%
$5.0 \times {10}^{5} {m}/{s}$
0%
$7.5 \times {10}^{5} {m}/{s}$
0%
$1.5 \times {10}^{6} {m}/{s}$
0%
$3.0 \times {10}^{6} {m}/{s}$
0%
$6.0 \times {10}^{6} {m}/{s}$

Explanation

The magnetic force does not change the speed of a charged particle, it only changes the velocity i.e. the direction of charged particle ,

The magnetic force on a charge

$q$ is given by

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

The acceleration by this force on of mass m of proton will be

$\vec{a}=q\left(\vec{v}\times\vec{B}\right)/m$

The speed will change only if any component of

$\vec{a}$ is in the direction of velocity but cross product shows that acceleration is perpendicular to velocity so can only change the direction of velocity and speed will remain same as

$1.5\times{10}^{6} m/s$

The diagram above shows a negatively charged ball moving to the right below a wire carrying conventional current to the right. Both are in the plane of the screen.
What is the direction of the force on the negatively charged ball at the place it is pictured?

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up toward the top of the screen
0%
down toward the bottom of the screen
0%
toward the left side of the screen
0%
toward the right side of the screen
0%
The force is zero, since the negative charge is moving parallel to the current.

Three different identical charge particles are pictured in the same magnetic field which points into the screen (represented by blue X's). The particles are moving at the same speeds but in different directions, as indicated by the red arrows.
How do the particles rank. In terms of the force they experience due to their movements in the magnetic field greatest first?

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1, 2, 3
0%
1 and 2 tie, 3
0%
3, 1 and 2 tie
0%
3, 2, 1
0%
all tie,

Explanation

The equation for the force on a charged particle moving through a magnetic field IS

$F=Bv\sin \theta$

where theta is the angle between the velocity direction and the direction of the magnetic field.If we look carefully at the directions here, we see that the velocity directions are all at ninety degrees to the magnetic field. All velocity directions are in the plane of the screen. The magnetic field direction is perpendicular to the screen.The force that all the charged particles experience is the same.

so correct Choice is option "E"

By which factor the magnetic field produced by a wire at distance 2 cm from the wire than at 4 cm from the wire is stronger if wire length is 2 m and carries a 10-amp current :

0%
2
0%
$2\sqrt{2}$
0%
4
0%
$4\sqrt{2}$
0%
8

Explanation

Magnetic field due to long wire is

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

As I is same for both cases so

$B\propto 1/r$

So,

$\dfrac{B_2}{B_1}=\dfrac{r_1}{r_2}=\dfrac{4}{2}=2$

A proton of charge

$e$ moving at speed

$v_0$ is placed midway between two parallel wires a distance

$a$ apart, each carrying current

$I$ in opposite directions.
The force on the proton is:

0%
$0$
0%
$ev_0\cfrac{_0I}{2\pi a}$
0%
$ev_0\cfrac{_0I}{2\pi (2a)}$
0%
$2ev_0\cfrac{_0I}{2\pi \frac{a}{2}}$
0%
Unable to be determined

A proton of charge

$'e'$ moving at speed

$'v_0\ '$ is placed midway between two parallel wires

$a$ distance a apart, each carrying current

$I$ in the same direction.
The force on the proton is

0%
$0$
0%
$ev_0\cfrac{_0I}{2\pi a}$
0%
$ev_0\cfrac{_0I}{2\pi (2a)}$
0%
$ev_0\cfrac{_0I}{2\pi \frac{a}{2}}$
0%
Unable to be determined

Explanation

Magnetic field due to a straight wire at a distance d

$B =\dfrac{\mu_o I}{2\pi d}$

Thus magnetic field at p due to wire 1

$B_1 = \dfrac{\mu_o I}{2\pi (a/2)} = \dfrac{\mu_o I}{\pi a}$ (into the paper)

Magnetic field at p due to wire 2

$B_2 = \dfrac{\mu_o I}{2\pi (a/2)} =\dfrac{\mu_o I}{\pi a}$ (out of the paper)

$\therefore$ Net magnetic field at p

$B_{net} =B_1 - B_2 = 0$

$\implies$ Magnetic force experienced by proton

$F = ev_o B_{net} = 0$ Newton

A wire carrying a current of 4 A is in a 0.5 T magnetic field as shown above.
If the direction of electron flow is to the right as shown, what is the magnitude and direction of the force (per unit length) on the wire?

0%
$2N/m$ into the page
0%
$8N/m$ out of the page
0%
$2N/m$ to the top of the page
0%
$2N/m$ to the bottom of the page
0%
$8N/m$ to the top of the page

Explanation

Given :

$\vec{B} = -0.5$

$\hat{z}$
Current flowing through the wire

$\vec{I} = - 4 \hat{x}$ A (because current flows in a direction opposite to that of electron flow)

Let the length of the wire be

$l$ .

Magnetic force

$\vec{F} = I(\vec{l} \times \vec{B}) = I lB[-\vec{x} \times (-\hat{z})]$

$\therefore$ Force per unit length

$\dfrac{\vec{F}}{l} =( 4 \times 0.5)$

$(-\hat{y}) =-2$

$\hat{y}$

$N/m$

Hence option D is correct.

A proton of charge

$e$ and mass

$m_p$ moves in a circular path of radius

$r$ in a uniform magnetic field

$B$ .
The momentum of the proton can be described by the expression:

0%
$eBr$
0%
$2eBr$
0%
$eBr^2$
0%
$eBrm_p$
0%
$2eBrm_p$

Explanation

Momentum of the proton

$P = m_pv$

Using

$eB r = m_pv$

$\implies$

$P = eBr$

A proton of charge

$e$ moving at speed

$v_0$ is placed midway between two parallel wires a distance

$a$ apart, each carrying current II in opposite directions.
The force on the proton is:

0%
$0$
0%
$ev_0\cfrac{\mu_0I}{2\pi a}$
0%
$ev_0\cfrac{\mu_0I}{2\pi (2a)}$
0%
$2ev_0\cfrac{\mu_0I}{2\pi \frac{a}{2}}$
0%
Unable to be determined

A particle of charge

$q$ and mass

$m$ moves in a circular path of radius

$r$ in a uniform magnetic field

$B$ .
The angular momentum of the particle can be described by the expression:

0%
$eBr$
0%
$2eBr$
0%
$eBr^2$
0%
$eBrm_p$
0%
$2eBrm_p$

Explanation

Angular momentum of the particle

$L = mv r$

Using

$eB r = mv$

$\implies$

$L = (eBr ) r =eBr^2$

A particle of charge

$q$ and mass

$m$ is moving at a speed

$v$ enters a uniform magnetic field of strength

$B$ as shown below.
How much work is done by the magnetic field on the charge as the field accelerates the charge into a circle of radius

$r$ ?

0%
$0$
0%
$\cfrac{1}{2}mv^2$
0%
$qvBr$
0%
$mv^2$
0%
Cannot be determined

Explanation

As the charge is moving perpendicular to the magnetic field, therefore, direction of force on charge by magnetic field will be given by Fleming's left hand rule, that is vertically downward (perpendicular to both velocity and magnetic field).

now work done is given by,

$W=Fd\cos\theta$ ,

but

$\theta=90^0$ , as force and displacement (velocity) are perpendicular to each other,

therefore,

$W=Fd\cos90^0=0$

A metallic ring of radius

$a$ and resistance

$R$ is held fixed with its axis along a spatially uniform magnetic field whose magnitude is

$B_{0}\sin (\omega t)$ . Neglect gravity. Then;

0%
The current in the ring oscillates with a frequency of $2\omega$
0%
The joule heating loss in the ring is proportional to $a^{2}$
0%
The force per unit length on the ring will be proportional to $B_{0}^{2}$
0%
The net force on the ring is non-zero

Explanation

According to Lenz's law, magnitude of induced current will be

$i=\dfrac{d\phi/dt}{R}$

where

$\phi=B.A$ where

$B = B_{0}\sin \omega t$ .

$i=\dfrac{B_0 \omega cos \omega t \pi a^2}{R}$

$P=\int i^2Rdt \propto a^4$

$F=idl B$

$F\propto B_0^2$

If the value of

$\theta$ increases then the magnetic moment value

0%
Increases
0%
Decreases
0%
Remains same
0%
Cannot be said

Explanation

Let l be length of wire, r be radius

$\theta$ be angle subtended at center.

Now, magnetic moment,

$M=m\times r$

Where m is pole strength for straight wire.

$M=m\times l$ (1)

Now when l is arc with angle then,

$r=\dfrac{l}{\theta}$

Now new magnetic moment,

$M^{'}=m\times l^{'}=m\times\dfrac{l}{\theta}=m$ and

$M\propto\dfrac{1}{\theta}$ . Thus decreases.

Find the resultant magnetic moment when

$\theta = {60}^{o}$ .

0%
${M}/{\pi}$
0%
${2M}/{\pi}$
0%
${3M}/{\pi}$
0%
${\pi}/{2M}$

Explanation

Let l be length of wire, r be radius

$\theta$ be angle subtended at center.

Now, magnetic moment,

$M=m\times r$

Where m is pole strength for straight wire.

$M=m\times l$ (1)

Now when l is arc then,

$r=\dfrac{l}{\theta}=\dfrac{l\times 3}{\pi}$

Now new magnetic moment,

$M^{'}=m\times l^{'}=m\times\dfrac{l_3}{\pi}[\because M=m\times l]$

$M^{'}=\dfrac{3M}{\pi}$

A curved magnet which occupies

$\dfrac{1}{6}th$ of a circle of magnetic moment

$\dfrac{30}{\pi} A {m}^{2}$ is made straight. Find final magnetic moment.

0%
$10 A {m}^{2}$
0%
$20 A {m}^{2}$
0%
$30 A {m}^{2}$
0%
$40 A {m}^{2}$

Explanation

Angle

$\theta=\dfrac{360}{6}$ =

$60°$

Length

$ACB=2R \sin { 30° } =R$

Length

$ADB=R\theta=\dfrac{R \pi}{3}$

When magnet is bent effective length is ACB .

Magnetic moment

$M=m(R)$ ..(1)

$m$ =pole strength

When magnet is straight effecetive length is ADB.

New magnetic moment is

$M'$

$M'=m(R\dfrac{\pi}{3})$ ..(2)

From (1) and (2) ,

$M'=M\times \dfrac{\pi}{3}$

$=\dfrac{30}{\pi} \times \dfrac{\pi}{3}=10Am^{2}$

Find the resultant magnetic moment when

$\theta = {240}^{o}$ .

0%
${M}/{4 \pi}$
0%
${2M}/{ \pi}$
0%
${3M}/{ \pi}$
0%
${3\sqrt{3}M}/{4 \pi}$

Explanation

Let pole strength of wire is m and length = l

then

$M=mL$

Length

$AB=2*cos30^{\circ}$

$L=2\pi r*\frac{240^{\circ}}{360^{\circ}}$

$L=\frac{4\pi r}{3}$

$\therefore$ magnetic moment

${M}'=m(AB)$

$=m*2*r*\frac{\sqrt{3}}{2}$

$=m\sqrt{3}r$

$=m\sqrt{3}*\frac{3L}{4\pi }$

$=\frac{3\sqrt{3}}{4\pi }(mL)$

$=\frac{3\sqrt{3}}{4\pi }M$

The value of

$\mu$ is

$4 \pi \times {10}^{-7} H {m}^{-1}$ .

0%
True
0%
False

Explanation

The physical constant

$μ_0$ commonly called the vacuum permeability, permeability of free space, ... In the reference medium of classical vacuum, μ0 has an exact defined value: .... The value of

$μ_0$ was chosen such that the rmks unit of current is equal in size to the ampere in the emu system: μ0 is defined to be

$4π × 10^{−7} H/m.$

A proton beam enters a magnetic field of

$10^{-4}Wb\ m^{-2}$ normally. If the specific charge of the proton is

$10^{11}C\ kg^{-1}$ and its velocity is

$10^{9}\ ms^{-1}$ , then the radius of the circle described will be

0%
$0.1\ m$
0%
$10\ m$
0%
$100\ m$
0%
$1\ m$

Explanation

Given :

$B = 10^{-4}$

$Wb/m^2$

$q = 10^{11}C/kg$

$v = 10^9 m/s$

Radius of circle described,

$r = \dfrac{mv}{Bq}$

$\therefore$

$r = \dfrac{10^9}{10^{-4} \times 10^{11}} = 100\ m$

A circular coil of radius 10 cm and 100 turns carries a current

$1 A$ . What is the magnetic moment of the coil?

0%
$3.142 \times 10^4 A m^2$
0%
$10^4 A m^2$
0%
$3.142 A m^2$
0%
$3 A m^2$

Explanation

Given :

$r = 10$ cm

$= 0.1$ m

$N = 100$ tuns

$I = 1A$

Area of the circular coil

$A = \pi r^2 = 3.142 (0.1)^2$

$m^2$

Magnetic moment

$\mu = NIA$

$\therefore$

$\mu = (100)(1)(3.142)(0.1)^2 = 3.142$

$Am^2$

Pick out the WRONG statement.

0%
The gain in the K.E. of the electron moving at right angles to the magnetic field is zero
0%
When an electron is shot at right angles to the electric field, it traces a parabolic path
0%
An electron moving in the direction of the electric field gains K.E.
0%
An electron at rest experiences no force in the magnetic field

A stream of electrons and protons are directed towards a narrow slit in a screen (see figure). The intervening region has a uniform electric field

$\overrightarrow{E}$ (vertically downwards) and a uniform magnetic field

$\overrightarrow{B}$ (out of the plane of the figure) as shown. Then

0%
Electrons and protons with speed $\dfrac{|\overrightarrow{E}|}{|\overrightarrow{B}|}$ will pass through the slit
0%
Protons with speed $\dfrac{|\overrightarrow{E}|}{|\overrightarrow{B}|}$ will pass through the slit, electrons of the same speed will not
0%
Neither electrons nor protons will go through the slit irrespective of their speed
0%
Electrons will always be deflected upwards irrespective of their speed

An electron enters an electric field having intensity

$\vec { E } =3\hat { i } +6\hat { j } +2\hat { k }$

$V{m}^{-1}$ and magnetic field having induction

$\vec { B } =2\hat { i } +3\hat { j } T$ with a velocity

$\vec { V } =2\vec { i } +3\vec { j }$

${ms}^{-1}$ . The magnitude of the force acting on the electron is (Given

$e=-1.6\times {10}^{-19}C$ )

0%
$2.02\times {10}^{-18}N$
0%
$5.16\times {10}^{-16}N$
0%
$3.72\times {10}^{-17}N$
0%
$4.41\times {10}^{-18}N$
0%
None of the above

Explanation

The electric force acting on the electron is

$qE=-e(3\hat{i}+6\hat{j}+2\hat{k})N$

The magnetic force acting on the electron=

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

$=q((2\hat{i}+3\hat{j})\times(2\hat{i}+3\hat{j}))$

$=0$

Hence net force acting on the electron is

$-e(3\hat{i}+6\hat{j}+2\hat{k})$

Magnitude of this force=

$e(\sqrt{3^2+6^2+2^2})$

$=7e=11.2\times 10^{-19}N$

A wire of length

$1 m$ is made into a circular loop and it carries a current of

$3.14 A$ . The magnetic dipole moment of the current loop (in

${Am}^{2}$ ) is :

0%
$1$
0%
$0.5$
0%
$0.25$
0%
$0.314$

Explanation

Given:

$L = 1\ m$ and

$i = 3.14A$

$= \pi A$

Radius of the circular loop,

$r =\dfrac{L}{2\pi}$

$\therefore$

$r = \dfrac{1}{2\pi}$

$m$

Area of the loop,

$A = \pi r^2 = \pi \times \dfrac{1}{4\pi^2} = \dfrac{1}{4\pi}$

$m^2$

Dipole moment,

$\mu = iA$

$\therefore$

$\mu = \pi \times \dfrac{1}{4\pi} = 0.25$

$Am^2$

The Biot Savart's Law in vector form is

0%
$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {dl(\vec {l}\times \vec {r})}{r^{3}}$
0%
$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {dl}\times \vec {r})}{r^{3}}$
0%
$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {r}\times \vec {dl})}{r^{3}}$
0%
$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {dl}\times \vec {r})}{r^{2}}$

Explanation

The Biot Savart's Law we know is given by:

$\vec{\delta B} = \dfrac{\mu_0}{4\pi}\dfrac{I dl \sin(\theta)}{r^2}\hat{r}$ , where

$\theta$ is the angle between the line element

$\vec{dl}$ and the radial unit vector

$\hat{r}$ .

Now we know:

$\hat{r} = \dfrac{\vec{r}}{r}$ , using this and the cross product of two vectors we get;

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

A circular coil of

$200$ turns and radius

$10 cm$ is placed in an uniform magnetic field of

$0.1 T$ normal to the plane of the coil. The coil carries a current of

$5 A$ . The coil is made up of copper wire of cross-sectional area

${10}^{-5}{m}^{2}$ and the number of free electrons per unit volume of copper is

${10}^{29}$ . The average force experienced by an electron in the coil due to magnetic field is

0%
$5 \times {10}^{-25}N$
0%
Zero
0%
$8 \times {10}^{-24}N$
0%
None of these

Explanation

$V_d=\dfrac{i}{nqA}$ ,

Force=

$F=q(V_d\times B)$ ,

$F= \dfrac{iB}{nA}=\dfrac{5\times0.1}{10^{29}\times10^{-5}}=5\times10^{-25}N$

If the work done in turning a magnet of magnetic moment

$M$ by an angle of

$90^o$ from the magnetic meridian is n times the corresponding work done to turn it through an angle of

$60^o$ , then the value of

$n$ is

0%
1
0%
2
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$

Explanation

Given,

Magnetic moment of magnet

$\theta_1= 90^0$

$\theta_2=60^0$

Work done in these process

We have,

$W=-MB(\cos {\theta}_2-\cos {\theta}_1)$
So,

$W_1=-MB(\cos 90^o-\cos 0^o) = MB$
and

$W_2=-MB(\cos 60^o-\cos 0^o) = \dfrac{1}{2}MB$
As

$W_1 = nW_2$

$\therefore n = \dfrac{W_1}{W_2}=\dfrac{MB}{\dfrac{1}{2}MB} = 2$

So the value of

$n=2$

A magnetic wire of dipole moment

$4\pi$

$Am^2$ is bent in the form of semi-circle. The new magnetic moment is?

0%
$4\pi$ $Am^2$
0%
$8\pi$ $Am^2$
0%
$4$ $Am^2$
0%
$8\ Am^2$

Explanation

Given:

The dipole moment of the wire is

$4\pi\ Am^2$ .

The Dipole moment of the wire is represented as:

$M_i=mL$

$m=\dfrac{4\pi}{L}$

Now, when the wire becomes a semicircle, the new magnetic moment will be:

$M_f=mL'$

The new distance between two poles is:

$L'=2r\Rightarrow2\dfrac{L}{\pi}$

So, new magnetic moment will be:

$M_f=m\times\dfrac{2L}{\pi}$

$M_f=\dfrac{4\pi}{L}\times\dfrac{2L}{\pi}$

$=8\ Am^2$

A current carrying loop is placed in a uniform magnetic field in four different orientations I, II, III and IV as shown in figure. Arrange them in decreasing order of potential energy.

0%
I > III > II > IV
0%
I > II > III > IV
0%
I > IV > II > III
0%
III > IV > I > II

Explanation

$\textbf{Hint: Use the formula of the potential energy of a current-carrying loop placed in a uniform magnetic field}$

$\textbf{Step1: Formula of potential energy}$

As we know that, potential energy of a magnet in a magnetic field

$U=-m\cdot B$

$=-mB\cos { \theta }$
where,

$m=$ magnetic dipole moment of the magnet

$B=$ magnetic field

$\textbf{Step2: Calculate potential energy in different cases}$
Case I

$\theta ={ 180 }^{ o }$

$\therefore { U }_{ 1 }=-mB\cdot \cos { { 180 }^{ o } }$

$=mB$

$\left[ \because \cos { { 180 }^{ o } } =-1 \right]$
Case II

$\theta ={ 90 }^{ o }$

${ U }_{ 2 }=0$

$\left[ \because \cos { { 90 }^{ o } } =0 \right]$
Case III

$\theta$ is acute angle

$\theta \in \left( 0,{ 90 }^{ o } \right)$

$\therefore \cos { \theta } =$ positive
Thus,

${ U }_{ 3 }=$ negative
Case IV

$\theta$ is obtuse

$\theta \in \left( { 90 }^{ o },{ 180 }^{ o } \right)$

$\therefore \cos { \theta } \in \left( 0,-1 \right)$
Thus,

${ U }_{ 4 }=$ positive
Therefore, decreasing order of PE is
I > IV > II > III

$\textbf{Hence option C correct}$

Two long parallel wires separated by

$0.1m$ carry currents if

$1A$ and

$2A$ respectively in opposite directions. A third current-carrying wire parallel to both them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of

0%
$0.5m$ from the 1st wire, towards the 2nd wire
0%
$0.2m$ from the 1st wire, towards the 2nd wire
0%
$0.1m$ from the 1st wire away from the 2nd wire
0%
$0.2m$ from the 1st wire, away from the 2nd wire

Explanation

Force due to current carrying wire on another wire is given by

$\dfrac{\mu_0 I_1I_2}{2 \pi r}$

Suppose wire is placed at a distance

$x$ at left side of first wire carrying current

$1 A$

Let's assume new wire is carrying current

$I$ A

Then force on new wire due to 1st wire

$= \dfrac{\mu_0 \times 1 \times I}{2\pi x}$ ...............(1)

and due to second wire

$= \dfrac{\mu_0 I \times 2}{2\pi(x+0.1)}$ ............(2)

Since both wires are carrying current in different direction, there force on new wire will be in opposite direction

On equating (1) and (2)

$\dfrac{\mu_0 \times 1 \times I}{2 \pi \times x }= \dfrac{\mu_0 \times I \times 2}{2 \pi (x+0.1)}$

On solving, we get

$x = 0.1 m$

Two wires of same length are shaped into square and a circle. if they carry same current ratio of magnetic moment is

0%
$\displaystyle 2:\pi$
0%
$\displaystyle \pi :3$
0%
$\displaystyle \pi :4$
0%
$\displaystyle 1:\pi$

Explanation

$\textbf{Hint:}$ Magnetic moment of current carrying loop is

$M=NiA$ .

Where n is number of turns ,i is current and A is area of loop.

$\textbf{Step 1-Magnetic moment of square}$

If length of wire is

$l$ and side of square is

$a$ .

Length of wire =circumference of square.

$\therefore l=4a$

$\Rightarrow a=\dfrac{l}{4}$

So magnetic moment of square is

$M_s=ia^2$

$\Rightarrow M_s=i\dfrac{l^2}{16}$ ...(1)

$\textbf{Step 2-Magnetic moment of circle}$

If radius of circle is

$r$ then

$l=2\pi r$

$\Rightarrow r=\dfrac{l}{2\pi}$

So magnetic moment of circle is

$M_c=i\pi r^2$

$\Rightarrow M_c=i\pi\dfrac{l^2}{4\pi^2}$

$\Rightarrow M_c=i\dfrac{l^2}{4\pi}$ ......(2)

$\textbf{Step 3-ratio of both magnetic moments}$

Divide equation (1) by (2)

$\dfrac{M_s}{M_c}=\dfrac{\pi}{4}$

Hence the correct answer is (C).

Four very long current carrying wires in the same intersect to form a square

$40.0\ cm$ on each side as shown in the figure. What is the magnitude of current

$I$ so that the magnetic field at the centre of the square is zero?

0%
$22A$
0%
$38A$
0%
$2A$
0%
$18A$

Explanation

Magnitude of the magnetic field
At a perpendicular distance

$x$ from infinite long wire is given by

$B = \dfrac {\mu d}{2\pi x}$
For wire

$AB \ B_{1} \dfrac {20\mu_{0}}{2\pi x}\hat {k} .... (i)$
Due to

$BC$

$B_{2} = \dfrac {\mu_{0}l}{2\pi x}(-\hat {k}) .... (ii)$
Due to

$DC$

$B_{3} = \dfrac {8\mu_{0}}{2\pi x}(-\hat {k}) ...,. (iii)$
Due to

$AD$

$B_{4} = \dfrac {10\mu_{0}}{2\pi x} (-\hat {k}) .... (iv)$
From Eqs. (i), (ii), (iii) and (iv)

$B_{1} + B_{2} + B_{3} + B_{4} = 0$ [From question]

$\Rightarrow \dfrac {20\mu_{0}}{2\pi x} - \dfrac {\mu_{0}l}{2\pi x} = \dfrac {8\mu_{0}}{2\pi x} + \dfrac {10\mu_{0}}{2\pi x}$

$\Rightarrow 20 - l = 8 + 10$

$\Rightarrow l = 20 - 18$

$l = +2A$
(Positive sign shows the direction of current is same direction as that of given direction)

A wire of length

$1m$ is moving at a speed of

$2m{ s }^{ -1 }$ perpendicular to its length in a homogeneous magnetic field of

$0.5T$ . If the ends of the wire are joined to a circuit of resistance the

$6 \Omega$ , then the rate at which work is being done to keep the wire moving at constant speed is

0%
$1W$
0%
$\cfrac { 1 }{ 3 } W$
0%
$\cfrac { 1 }{ 6 } W$
0%
$\cfrac { 1 }{ 12 } W$

Explanation

Rate of work

$\cfrac { W }{ t } =P$
But

$P=Fv$
Also,

$F=Bil=B\left( \cfrac { Bvl }{ R } \right) l$
[

$\because$ induced current

$i=\cfrac { Bvl }{ R }$ ]

$\therefore P=B\left( \cfrac { Bvl }{ R } \right) lv=\cfrac { { B }^{ 2 }{ v }^{ 2 }{ l }^{ 2 } }{ R } =\cfrac { { (0.5) }^{ 2 }\times { (2) }^{ 2 }\times { (1) }^{ 2 } }{ 6 } =\cfrac { 1 }{ 6 } W$

0.8 J work is done in rotating a magnet by

$60^o$ , placed parallel to a uniform magnetic field. How much work is done in rotating it

$30^o$ further?

0%
$0.8\times 10^7 erg$
0%
$0.8 erg$
0%
$8 J$
0%
$0.4 J$

Explanation

Work done,

$W=MB (\cos {\theta}_1 - \cos {\theta}_2)$
When the magnet is rotated from

$0^o$ to

$60^o$ , then work done is 0.8 J

$0.8 = MB(\cos 0^o - \cos 60^o) = \dfrac{MB}{2}$

$MB=0.8\times 2 = 1.6 N-m$
In order to rotate the magnet through an angle of

$30^o$ , i.e., from

$60^o$ to

$90^o$ , the work done is

$W' = MB(\cos 60^o - \cos 90^o)$

$=MB(\dfrac{1}{2}-0)=\dfrac{MB}{2}$

$W'=\dfrac{1.6}{2}=0.8 J$

$=0.8\times 10^7 erg$

A horizontal overhead power line carries a current of 90A in east to west direction. Magnitude of magnetic field due to the current 1.5 m below the line is :

0%
$1.2 T$
0%
$1.2 \times 10^{-10} T$
0%
$0T$
0%
$1.2 \times 10^{-5} T$

Explanation

Magnitude of magnetic field at a distance

$r$ due to straight current carrying conductor is given by

$B=\dfrac{\mu_0}{2\pi}\dfrac{i}{r}$

Given :

$i = 90$ A and

$r = 1.5$ m

$B=2\times 10^{-7} \times \dfrac{ 90}{1.5}=1.2\times 10^{-5}T$

Two parallel long wires carry currents

$i_{1}$ and

$i_{2}$ with

$i_{1} > i_{2}$ . When the currents are in the same direction, the
magnetic field midway between the wires is

$10 \mu T$ . When the direction of

$i_{2}$ is reversed, it becomes

$40 \mu T$ . The
ratio

$\dfrac {i_{1}}{i_{2}}$ is

0%
$3:4$
0%
$5:3$
0%
$7:11$
0%
$11:7$

Explanation

When the current in the wires is in same direction. Magnetic field at mid point

$O$ due to

$I$ and

$II$ wires are respectively

${ B }_{ I }=\cfrac { { \mu }_{ 0 } }{ 4\pi } \cfrac { 2{ i }_{ 1 } }{ x } \otimes ;\quad { B }_{ II }=\cfrac { { \mu }_{ 0 } }{ 4\pi } \cfrac { 2{ i }_{ 2 } }{ x } \oslash$
So, the net magnetic field at

$O$

${ B }_{ net }=\cfrac { { \mu }_{ 0 } }{ 4\pi } \times \cfrac { 2 }{ x } \left( { i }_{ 1 }-{ i }_{ 2 } \right)$

$\Rightarrow 10\times { 10 }^{ -6 }=\cfrac { { \mu }_{ 0 } }{ 4\pi } \cfrac { 2 }{ x } \left( { i }_{ 1 }-{ i }_{ 2 } \right) ...(i)$
when the direction of

${i}_{2}$ is reversed

${ B }_{ I }=\cfrac { { \mu }_{ 0 } }{ 4\pi } \cfrac { 2{ i }_{ 1 } }{ x } \otimes \quad ;{ B }_{ II }=\cfrac { { \mu }_{ 0 } }{ 4\pi } \cfrac { 2{ i }_{ 2 } }{ x } \otimes$
so, net magnetic field at

$O$

${ B }_{ net }=\cfrac { { \mu }_{ 0 } }{ 4\pi } \times \cfrac { 2 }{ x } \left( { i }_{ 1 }+{ i }_{ 2 } \right)$

$\Rightarrow 40\times { 10 }^{ -6 }=\cfrac { { \mu }_{ 0 } }{ 4\pi } \cfrac { 2 }{ x } \left( { i }_{ 1 }+{ i }_{ 2 } \right) ...(ii)$
Dividing EQ. (ii) by (i) we get

$\cfrac { { i }_{ 1 }+{ i }_{ 2 } }{ { i }_{ 1 }-{ i }_{ 2 } } =\cfrac { 4 }{ 1 } \Rightarrow \cfrac { { i }_{ 1 } }{ { i }_{ 2 } } =\cfrac { 5 }{ 3 }$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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