MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 9

A particle with certain charge enters a region of constant, uniform and mutually orthogonal fields E and B with a velocity u perpendicular to both E and B and comes out without any change in magnitude or direction u. This means that

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Force acting on particle is Inclined at an angle of $180^o$
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Force acting on particle is highly directional
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Force acting on charged particle is zero
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Charge on particle is too high

Explanation

Force acting on changed particle is zero, as v of changed particle remains constant

$F= q(v\times B)$

If flow of electric current is parallel to magnetic field, force will be:

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Negative
0%
Maximum
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Zero
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None of these

Explanation

Electric current is basically flow of +ve charge (assumed) and force on a charge moving in any direction with velocity

$\vartheta$ is given by

$F=q(\vec{\vartheta }*\vec{B})$

if electric current is parallel to

$\vec{B}$ ,then

$\vec{\vartheta }*\vec{B}=0$

$\therefore F=0$

A long straight wire carrying current

$I$ is bent at its mid-point to form an angle of

$45^{\circ}$ . The field at point

$P$ as shown in the figure is

0%
$\dfrac {\mu_{0}I}{4\pi R}(\sqrt {2} - 1)$
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$\dfrac {\mu_{0}I}{4\pi R}(\sqrt {2} + 1)$
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$\dfrac {\mu_{0}I}{4\sqrt {2}\pi R}(\sqrt {2} + 1)$
0%
$\dfrac {\mu_{0}I}{4\sqrt {2}\pi R}(\sqrt {2} - 1)$

Explanation

Since the point P lies on the axis of straight part OA and hence magnetic field due to this part will be zero.

$B=\frac{\mu _{0}I}{4\pi r}(sin \alpha +sin \beta)$

for this case

Since both ends O and B are on the same side of normal Pn,

therefore

$\phi _{1}=-45^{\circ}$ and

$\phi _{2}=90^{\circ}$

$\therefore B=\frac{\mu _{0}I}{4\pi d}(sin (-45^{\circ}) +sin (90^{\circ}))$

$=\dfrac{\mu _{0}I}{4\pi(Rcos45^{\circ})}[-sin45^{\circ}+1]$

$=\dfrac{\mu _{0}I}{4\pi R*\frac{1}{\sqrt{2}}}\left ( 1- \frac{1}{\sqrt{2}}\right )$

$=\dfrac{\mu _{0}I}{4\pi R}(\sqrt{2}-1)$

The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire is

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$\dfrac {\pi}{2\sqrt {2}}$
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$\dfrac {\pi}{4\sqrt {2}}$
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$\dfrac {\pi^{2}}{4\sqrt {2}}$
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$\dfrac {\pi^{2}}{8\sqrt {2}}$

Explanation

Magnetic field at the centre of a circular coil

$B_{Circular} = \dfrac {\mu_{0}}{4\pi} \dfrac {2\pi i}{r} = \dfrac {\mu_{0}}{4\pi} \dfrac {4\pi^{2}i}{L} \left [\because L = 2\pi r\ and\ r = \dfrac {1}{2\pi}\right ]$
Magnetic field at the centre of square coil

$B_{Square} = \dfrac {\mu_{0}}{4\pi}\dfrac {8\sqrt {2}i}{a}= \dfrac {\mu_{0}}{4\pi} \dfrac {32\sqrt {2}i}{L}\left [\because L = 4a\ and\ a = \dfrac {L}{4}\right ]$
Hence,

$\dfrac {B_{Circular}}{B_{Square}} = \dfrac {\dfrac {mu_{0}}{4\pi}\dfrac {4\pi^{2}i}{L}}{\dfrac {\mu_{0}}{4\pi}\dfrac {32\sqrt {2}i}{L}} = \dfrac {\pi^{2}}{8\sqrt {2}}$ .

Potential energy of a bar magnet, of magnetic moment M is placed in a magnetic filed of induction B such that it makes an angle

$\theta$ with the direction of B is?

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$MB\,sin\, \theta$
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$-MB\,cos\, \theta$
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$MB(1-cos\theta)$
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$MB(1+cos\theta)$

Explanation

$dw=-\tau .d\theta =-MB\sin \theta .d\theta$ where

$\theta$ is angle between

$\vec{M}$ and

$\vec{B}$ .

Since work done

$=$ potential energy lost

$=>du=-dw=MB\sin \theta .d\theta$

$=>u=\int MB\sin \theta .d\theta =-MB\cos \theta$

In the given figure, what is the magnetic field induction at point O.

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$\displaystyle\frac{\mu_0l}{4\pi r}$
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$\displaystyle\frac{\mu_0l}{4r}+\frac{\mu_0l}{2\pi r}$
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$\displaystyle\frac{\mu_ol}{4r}+\frac{\mu_0l}{4\pi r}$
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$\displaystyle\frac{\mu_0l}{4r}-\frac{\mu_0l}{4\pi r}$

Explanation

Magnetic field due to straight wire above O is zero
i.e.,

$B_1=0$ (since,

$\theta =0^o$ )
The magnetic field due to semi circular part

$B_2=\left(\displaystyle\frac{\mu_0nI}{2r}\right)=\frac{\mu_0(1/2)I}{2r}=\frac{\mu_0I}{4r}$
The magnetic field due to lower straight portion

$B_3=\displaystyle\frac{1}{2}\frac{\mu_0I}{2\pi r}$ (upward)
Net magnetic field

$B=B_1+B_2+B_3$

$=0+\displaystyle\frac{\mu_0I}{4r}+\frac{\mu_0I}{4\pi r}$ (upwards).

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is

$60^0$ and one of the fields has a magnitude of

$1.2 \times 10^{-2}$ T. If the dipole comes to stable equilibrium at an angle of

$30^0$ with this field, then the magnitude of the field is

0%
$1.2 \times 10^{-4}$ T
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$2.4 \times 10^{-4}$ T
0%
$1.2 \times 10^{-2}$ T
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$2.4 \times 10^{-2}$ T

Explanation

Here,

$\theta = 60^0, B_1 = 1.2 \times 10^{-2}T$

$\theta_1 = 30^0 and \theta_2 = 60^0 - 30^0 = 30^0$ in stable equilibrium,

torques due to two fields must be balanced i.e.

$T_1 = T_2$

$\Rightarrow MB_1 sin \theta_1 = MB_2 sin \theta_2$

$B_2 = B_1 \dfrac{sin \theta_1}{sin \theta_2} = B_2 \dfrac{sin 30^0}{sin 30^0} = B_1 = 1.2 \times 10^{-2}T$

1085066_943043_ans_35cdd384c40d4866bccdcb0fb3278853.jpg

Assertion : Diamagnetism is universal, it is present in all materials.
Reason : Field due to induced magnetic moment is opposite to the magnetising field

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If both assertion and reason are true and reason is the correct explanation of assertion.
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If both assertion and reason are true and reason is not the correct explanation of assertion.
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If assertion is true but reason is false.
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If both assertion and reason are false

In an inertial frame of reference, the magnetic force on a moving charged particle is

$\bar{F}$ . Its value in another internal frame of reference will be

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remain same
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changed due to change in the amount of charge
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changed due to change in velocity of charged particle
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changed due to change in field direction

Explanation

$\overset{\rightarrow}{F} = \overset{\rightarrow}q ( \overset{\rightarrow}{v} \times \overset{\rightarrow}{B}) \,\, \,\, \therefore F = qv B sin \theta$
which shows magnetic force is velocity dependent due to which it differs from one inertial frame to another.

Imagine yourself to be swimming in the wire in the direction of the current and facing a magnetic needle, then the north pole of the needle is deflected towards your left hand. Which is this rule?

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Ampere's rule
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Faraday right hand rule
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Faraday left hand rule
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Maxwell's rule

Two free parallel wires carrying currents in opposite directions.

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Do not affect each other
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Attract each other
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Repel each other
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None of these

A short bar magnet placed with its axis at

$30^0$ with a uniform external magnetic field of

$0.35T$ experiences a torque of magnitude equal to

$4.5 \times 10^{-2}$ N m. The magnitude of magnetic moment of the given magnet is

0%
$26 JT^{-1}$
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$2.6 JT^{-1}$
0%
$0.26 JT^{-1}$
0%
$0.026 JT^{-1}$

Explanation

Here,

$\theta = 30^0, B = 0.35T$

$T = 4.5 \times 10^{-8}Nm$

then,

$m = \dfrac{T}{B sin \theta} = \dfrac{4.5 \times 10^{-2}}{0.35 \times sin 30^0}$

$m = \dfrac{4.5 \times 10^{-2}}{0.35 \times \dfrac{1}{2}} = \dfrac{2 \times 4.5 \times 10^{-2}}{0.35} = 0.26 J T^{-1}$

A uniform conduction wire of length

$10 \,g$ and resistance

$R$ is wound up into four turns as a current carrying coil in the shape of equilateral triangle of side

$a$ . If current

$I$ is flowing through the coil then the magnetic moment of the coil is

0%
$\dfrac{\sqrt3}{2}a^2I$
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$\dfrac{a^2I}{\sqrt3}$
0%
$\sqrt3a^2I$
0%
$\dfrac{2a^2I}{\sqrt3}$

Explanation

It is given that

Number of turns $n=4$

Length of the uniform conducting wire $l=10\,a$

Area of an equilateral triangle $A=\dfrac{\sqrt{3}}{4}{{a}^{2}}$ where a is the side of triangle

Magnetic moment

$M=nIA$

$M=4\times I\times \dfrac{\sqrt{3}}{4}{{(10)}^{2}}$

$M=\sqrt{3}{{a}^{2}}I$

Ampere's circuital law is given by:

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$\oint \bar{H} . \bar{dl} = \mu _0 I_{enc}$
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$\oint \bar{B} . \bar{dl} = \mu _0 I_{enc}$
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$\oint \bar{B} . \bar{dl} = \mu _0 I$
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$\oint \bar{H} . \bar{dl} = \mu _0 I$

Explanation

The line integral of the magnetic field of induction

$\vec B$ around any closed path in free space is equal to absolute permeability of free space

$μ_o$ times the total current flowing through area bounded by the path.

Ampere's circuital law is given by:

$\oint \vec B.d\vec l=\mu_0 I_{enc}$

The correct option is B.

1498487_943869_ans_01cb87a335e242d7ab13fad6c11a2fb9.PNG

A circular coil of magnetic moment

$0.355$ J

$T^{-1}$ rests with its plane normal to an external field of magnitude

$5.0\times 10^{-2}$ T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of

$2$ Hz. The moment of inertia of the coil about its axis of rotation is?

0%
$1.13\times 10^{-1}$ kg $m^2$
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$1.13\times 10^{-2}$ kg $m^2$
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$1.13\times 10^{-3}$ kg $m^2$
0%
$1.13\times 10^{-4}$ kg $m^2$

Explanation

Here,

$M=0.355$ J

$T^{-1}$

$B=0.5\times 10^{-2}$ T, v

$=2$ Hz
As

$v=\dfrac{1}{2\pi}\sqrt{\dfrac{MB}{I}}$

$\therefore v^2=\dfrac{1}{4\pi^2}\dfrac{MB}{I}$

$\Rightarrow I=\dfrac{MB}{4\pi^2v^2}=\dfrac{0.355\times 5\times 10^{-2}}{4\times (3.14)^2\times 2^2}=\dfrac{1775\times 10^{-2}}{157.75}$

$=1.13\times 10^{-4}$ kg

$m^2$

Amperes circuital law is given by

0%
$\oint\bar H. \bar {dl} =\mu_0 I_{enc}$
0%
$\oint\bar B. \bar {dl} =\mu_0 I$
0%
$\oint\bar B. \bar {dl}=\mu_0 J$
0%
$\oint\bar H. \bar {dl}=\mu_0 J$

Explanation

Ampere 's circuital law states that relationship between thecurrent and the magnetic field

$\phi\overrightarrow { B } \overrightarrow { dl } =\mu_0 I$

B is correct option

An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 X

$10^{-19}$ C,

$m_e$ = 9.1 X

$10^{-31}$ kg)

0%
2.58 X $10^{-4}$ m
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3.58 X $10^{-4}$ m
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2.58 X $10^{-3}$ m
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3.58 X $10^{-3}$ m

Explanation

$\displaystyle E = \frac{1}{2} mv^2 \Rightarrow v = \sqrt{ \dfrac{2E}{m} }$

$\displaystyle r = \frac{mv}{Be} = \frac{m}{Be} \sqrt{ \dfrac{2E}{m} } = \dfrac{\sqrt{2mE}}{Be}$

$\displaystyle r = \frac{\sqrt{2 \times 1800 \times 1.6 \times 10^{-19} \times 9.1 \times 10^{-31}}}{1.6 \times 10^{-19} \times 0.4} = 3.58 \times 10^{-4} m$

The magnetic induction at a point

$1$

$\overset{o}{A}$ away from a proton measured along its axis of spin is (magnetic moment of the proton is

$1.4\times 10^{-26}$ A

$m^2$ ).

0%
$0.28$ mT
0%
$28$ mT
0%
$0.028$ mT
0%
$2.8$ mT

Explanation

On the axis of a magnetic dipoles, magnetic induction is

$B=\dfrac{\mu_o}{4\pi}\times \dfrac{2m}{r^3}$

Here,

$r=1$

$\overset{o}{A}=10^{-10}$ m,

$m=1.4\times 10^{-26}$ A

$m^{-2}$ and

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

$A^{-2}$

$\therefore B=\dfrac{10^{-7}\times 2\times 1.4\times 10^{-26}}{(10^{-10})^3}=\dfrac{10^{-7}\times 2.8\times 10^{-26}}{10^{-30}}$

$=2.8\times 10^{-3}$ T

$=2.8$ mT.

A 200 turn closely wound circular coil of radius 15 cm carries a current of 4 A. The magnetic moment of this coil is:

0%
$36.5 A\,m^2$
0%
$56.5 A\,m^2$
0%
$66.5A\,m^2$
0%
$108A\,m^2$

Explanation

The magnetic moment is given by

$| \bar m| = NIA = NI \pi r^2$

$= 200 \times 4 \times 3.14 \times (15 \times 10^{-2})^2$

$= 200 \times 4\times 3.14 \times 15 \times 15 \times 10^{-4}$

$= 56.5 A m^2$

If an electron is moving with velocity

$\bar{v}$ produces a magnetic field

$\bar{B}$ , then

0%
the direction of field $\bar{B}$ will be same as the direction of velocity $\bar{v}$
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the direction of field $\bar{B}$ will be opposite as the direction of velocity $\bar{v}$
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the direction of field $\bar{B}$ will be perpendicular as the direction of velocity $\bar{v}$
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the direction of field $\bar{B}$ does not depend upon the direction of velocity $\bar{v}$

Explanation

According to Biot-Savart's law, the magnetic field

$\displaystyle \overset{\rightarrow}{B} = \frac{\mu_o}{4 \pi} . \frac{q (\overset{\rightarrow}{v} \times \overset{\rightarrow}{r} ) }{r^3}$
The direction of

$\overset{\rightarrow}{B}$ will be along

$\overset{\rightarrow}{v} \times \overset{\rightarrow}{r}$ i.e. perpendicular to the plane containing

$\overset{\rightarrow}{v}$ and

$\overset{\rightarrow}{r}$ .

The torque and magnetic potential energy of a magnetic dipole in most stable position in a uniform magnetic field(

$\bar{B}$ ) having magnetic moment (

$\bar{m}$ ) will be.

0%
$-$ mB, zero
0%
mB, zero
0%
Zero, mB
0%
Zero, $-$ mB

Explanation

Torque,

$\bar{\tau}=\vec{m}\times \vec{B}=mB\sin\theta$
and magnetic potential energy

$U_m=\vec{m}\cdot \vec{B}=mB\cos \theta$
at

$\theta =0^o$ the dipole will be in most stable position

$\tau =mB\sin\theta =mB\sin 0^o=0$
and

$U_m=-mB\cos\theta =-mB\cos \theta^o=-mB$

Two charged particles traverse identical helical j paths in a completely opposite sense in a uniform magnetic field

$\bar B=B_0\hat {k}$ .

0%
They have equal z-components of momenta.
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They must have equal charges.
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They necessarily represent a particle-antiparticle pair.
0%
The charge to mass ratio satisfy:

${\left (\dfrac{e}{m} \right )}_1+{\left (\dfrac{e}{m} \right )}_2=0.$

Explanation

In this situation if the particle is thrown in x-y plane (as shown in figure) at some angle θ with velocity v, then we have to resolve the velocity of the particle in rectangular components, such that one component is along the field (v cosθ) and other one is perpendicular to the field (v sinθ). We find that the particle moves with constant velocity v cosθ along the field. The distance covered by the particle along the magnetic field is called pitch.

The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by

$p = T ( v \cos \theta ) = 2 p \dfrac { m } { q B } ( v \cos \theta )$

For given pitch $p$ correspond to charge particle, we have

$\quad \dfrac { q } { m } = \dfrac { 2 \pi v \cos \theta } { q B } =$ constant

Here in this case, charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B, LHS for two particles should be same and of opposite sign.

$\left( \dfrac { e } { m } \right) _ { 1 } + \left( \dfrac { e } { m } \right) _ { 2 } = 0$

1058703_949961_ans_b834b35f0c7742feaf3133404d9fef05.png

Similar pole each of pole strength m are placed at a distance of

$1, 2, 4, 8$ ,_________ meters from the origin on the x-axis. Where do you place a similar pole on the other side of the origin so that the origin becomes a neutral point.

0%
$0.5$ m
0%
$0.5774$ m
0%
$0.866$ m
0%
$1$ m

Explanation

$\dfrac{\mu_op}{4\pi x^2}=\dfrac{\mu_0p}{4\pi \times 1^2}+\dfrac{\mu_0\times p}{4\pi \times 2^2}+\dfrac{\mu_0p}{4\pi\times 4^2}+$ ________

$\dfrac{1}{x^2}=1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}$ _______

$=\dfrac{1}{1-\dfrac{1}{4}}=\dfrac{4}{3}$

$x=\dfrac{\sqrt{3}}{2}=0.866$ m

The magnetic field at the centre of a circular loop of area A is B. The magnetic moment of the loop is

0%
$\dfrac{BA^2}{\mu_0\pi}$
0%
$\dfrac{BA\sqrt A}{\mu}$
0%
$\dfrac{BA\sqrt A}{\mu_0\pi}$
0%
$\dfrac{2BA\sqrt A}{\mu_0\sqrt\pi}$

Explanation

Let r be the radius of the circular loop.

$\therefore$ A =

$\displaystyle \pi r^2$ or

$r$ =

$\displaystyle \sqrt{\frac{A}{\pi}}$
Magnetic field at the centre of the loop is

$B$ =

$\dfrac{\mu_0I}{2r}=\dfrac{\mu_0I}{2\sqrt{\dfrac{A}{\pi}}}$ or

$I=\dfrac{2B}{\mu_0}\sqrt{\dfrac{A}{\pi}}$
The magnetic moment of the loop is
M=

$IA$ =

$\dfrac{2B}{\mu_0}\sqrt{\dfrac{A}{\pi}}A$ =

$\dfrac{2BA\sqrt A}{\mu_0\sqrt\pi}$

The magnetic moment of a short bar magnet placed with its magnetic axis at

$30^o$ to an external field of

$900$ G and experiences a torque of

$0.02$ N m is?

0%
$0.35$ A $m^2$
0%
$0.44$ A $m^2$
0%
$2.45$ A $m^2$
0%
$1.5$ A $m^2$

Explanation

Here,

$B=900$ Gauss

$=900\times 10^{-4}$ T.

$=9\times 10^{-2}$ T

$\tau =0.02$ N m and

$\theta =30^o$

$\because \tau =mB\sin \theta$

$\Rightarrow 0.02=m\times 9\times 10^{-2}\times \sin 30^o$

$0.02=9\times 10^{-2}\times \dfrac{1}{2}\times m$

$m=\dfrac{0.02\times 2}{9\times 10^{-2}}=0.44$ A

$m^2$ .

A straight steel wire of length

$l$ has magnetic moment

$m$ . If the wire is bent in the form of a semicircle, the new value of the magnetic dipole moment is _______

0%
$m$
0%
$\cfrac { m }{ 2 }$
0%
$\cfrac { m }{ \pi }$
0%
$\cfrac { 2m }{ \pi }$

Explanation

Length

$l$ , long wire is twisted in the state of half circle of span

$r$ , at that point, length of the wire is calculated as follows:

Circuit of the half circle (i.e.,

$l=\pi r$ )

$\Longrightarrow r=\cfrac { l }{ \pi }$

Now, attractive dipole=attractive dipole minute

$\times$ least detachment between two dipoles

Least division between the dipoles is=

$2r=\cfrac{2l}{\pi}$

$\therefore$ Attractive dipole=

$M \times \cfrac{2l}{\pi}=\cfrac{2m}{\pi}$

The magnetic moment associated with a circular coil of 35 turns and radius 25 cm, if it carries a current of 11 A is:

0%
$72.2 A\,m^2$
0%
$70.5 A\,m^2$
0%
$74.56A\,m2$
0%
$75.56A\,m^2$

Explanation

Given

$N = 35, \\r = 25 cm = 25 \times 10^{-2}m, \\ I = 11 A$

Then magnetic moment associated with this circular coil

$M = NIA = NI \pi r^2 = 35\times 11 \times 3.14 \times ( 25 \times 10^{-2})^2 = 75.56 A m^2$

A circular coil of

$300$ turns and diameter

$14$ cm carries a current of

$15$ A. The magnitude of magnetic moment associated with the loop is?

0%
$51.7$ J $T^{-1}$
0%
$69.2$ J $T^{-1}$
0%
$38.6$ J $T^{-1}$
0%
$19.5$ J $T^{-1}$

Explanation

Given,

$D=14cm$

$r=\dfrac{14}{2}=7cm=0.07m$

$N=300$

$I=15A$

The magnitude of the magnetic moment associate with the loop is

$M=NIA=NI\pi r^2$

$M=300\times 15\times 3.14\times 0.07\times 0.07$

$M=69.2Am^2$

The correct option is B.

Biot-Savart law indicates that the moving electrons (velocity

$\bar v$ ) produce a magnetic field

$\bar B$ such that:

0%
$\bar B \perp \bar v$
0%
$\bar B \parallel \bar v$
0%
it obeys inverse cube law.
0%
it is along the line joining the electron and point of observation.

Explanation

Magnetic field produced by charges moving with velocity

$\bar v$ , at a distance r is

$\bar B$ =

$\left ( \dfrac{\mu_0}{4\pi } \right )$ .q

$\dfrac{\bar v \times \hat r}{r^2}$

Therefore

$\bar B \perp \bar v$

A star shaped loop (with l = length of each edge)then the magnetic field at the centroid of the loop is:

0%
$\frac{3\mu_0i}{\pi l}$
0%
$\frac{3\mu_0i}{2\pi l}$
0%
$(3 -\sqrt3)\frac{\mu_0i}{\pi l}$
0%
$(3 +\sqrt3)\frac{3\mu_0i}{\pi l}$

Explanation

For each of the 12 wires, perpendicular distance,

$\alpha$ and

$\beta$ are the same.
B=

$\dfrac{\mu_0 i}{4\pi \dfrac{l\sqrt3}{2}}[sin\alpha + sin\beta]$

$\alpha =- 30^0$

$\beta =+60^0$
B=

$\dfrac{\mu_0i}{2\pi l \sqrt3}\Big[ \dfrac{\sqrt3}{2}- \dfrac{1}{2}\Big]\times 12$

$=\dfrac{\mu_0 i}{\pi l} ( 3- \sqrt3)$

Fill in the blanks
A charge is moving through a magnetic field. The force acting on the charge is maximum when the angle between the velocity of charge and the magnetic field is ______

0%
$\pi$
0%
$\pi/3$
0%
$\pi/2$
0%
$\pi/4$

Explanation

We know,

$\overrightarrow{F_B}=q\overrightarrow{v}\times\overrightarrow B=qvB\sin\theta$

Where:

$\theta$ = angle between v and B ,

Now for

$\overrightarrow F$ to be maximum,

$\sin\theta$ should be max,

i.e

$\theta =\dfrac{\pi}{2}$

Option

$\textbf C$ is the correct answer.

A magnetised wire of magnetic movement M and length l is bent in the form of a semicircle of radius r. The new magnetic moment is:

0%
M
0%
$\frac{2M}{\pi}$
0%
$\frac{M}{\pi}$
0%
None of the above

Explanation

After bending, distance between poles=

$2r$
Earlier distance =

$l(\pi r)$

$\dfrac{M'}{M} =\dfrac{m\times 2r}{m\times \pi r}=\dfrac{2}{\pi}$

$\therefore M'= M\times \dfrac{2}{\pi}$

The incorrect statement of the following is:

0%
Work done on a moving charge in a magnetic field is zero
0%
The kinetic energy of a moving charge in magnetic field is conserved
0%
Magnetic field is not produced due to electron moving in a straight line
0%
Kinetic energy of a moving charge in an electric field changes

Explanation

Movement of charge in a straight line results in a magnetic field being generated from the relation

$F=qvB\sin\theta$

A conducting square loop of side

$L$ and resistance

$R$ moves in its plane with a uniform velocity

$v$ perpendicular to one of its sides. A magnetic induction

$B$ , constant in time and space, pointing perpendicular and into the plane of the loop exists everywhere, then the current induced in the loop is

0%
$BLv/R$ clockwise
0%
$BLv/R$ anticlockwise
0%
$2 BLv/R$ anticlockwise
0%
$0$

Explanation

Polarity induced

$X$ is shown in fig.

Due to same voltage on same side no current flows.

1171316_1011215_ans_8d20d82d1ffd410cb347804c4ee1a70a.png

A proton and an alpha particle are separately projected in a region , where a uniform magnetic field exists. The initial velocities are perpendicular to the direction of magnetic field . If both the particles move along circles of equal radii , the ratio of momentum of proton to alpha particle(

$\dfrac{P_p}{P_a}$ )

0%
1
0%
1/2
0%
2
0%
1/4

A charged particle moves in a uniform magnetic field perpendicular to it, with a radius of curvature 4 cm. On passing through a metallic sheet it looses half of its kinetic energy, then the radius of curvature of the particle is :

0%
2 cm
0%
4 cm
0%
8 cm
0%
$2\sqrt2$ cm

Explanation

$R= \dfrac{mv}{qB} =\dfrac{m}{qB}\sqrt{\dfrac{2(KE)}{m}}= \dfrac{\sqrt{2m}}{qB}\sqrt{KE}$

$\Rightarrow \dfrac{R_1}{R_2}= \dfrac{\sqrt{KE}}{\sqrt{KE/2}}$

$\Rightarrow R_2= \dfrac{R_1}{\sqrt2}=\dfrac{4}{\sqrt2}= 2\sqrt2cm$

Potential energy of a bar magnet of magnetic moment

$M$ placed in a magnetic field of induction

$B$ Such that it makes an angle

$\theta$ with the direction of

$B$ is (take

$\theta={90}^{o}$ as datum)

0%
$-M\ B \sin {\theta}$
0%
$-M\ B \cos {\theta}$
0%
$M\ B (1-\cos {\theta})$
0%
$M\ B (1+\cos {\theta})$

Explanation

$dw=-\tau.d\theta=-MB\sin\theta.d\theta$ where

$\theta$ is angle between

$\vec{M}$ and

$\vec{B}$ .

Since work done=Potential energy lost

$\implies du=-dw=MB\sin\theta.d\theta$

$u=\int MB\sin\theta.d\theta=-MB\cos\theta$

Figure shows a square loop of side

$1m$ and resistance

$1\Omega$ . The magnetic field on left side of line PQ has a magnitude

$B=1.0T$ . The work done in pulling the loop out of the field uniformly in

$1s$ is

0%
$1J$
0%
$10J$
0%
$0.1J$
0%
$100J$

Explanation

$V=\dfrac l{t}=1m/s$

$W=Power\times Time$

$=\dfrac{(Blv)^2}{R}\times t$

$=\dfrac{B^2l^2}{Rt}=\dfrac{1^2\times 1^4}{1\times 1}=1J$

Figure shows a square current-carrying loop

$ABCD$ of side

$2\ m$ and current

$I=\dfrac { 1 }{ 2 } A$ . The magnetic moment

$\vec { M }$ of the loop is:

0%
$\left( \hat { i } -\sqrt { 3 } \hat { k } \right) A-{ m }^{ 2 }$
0%
$\left( \hat { j } -\hat { k } \right) A-{ m }^{ 2 }$
0%
$\left( \sqrt { 3 } \hat { i } -\hat { k } \right) A-{ m }^{ 2 }$
0%
$\left( \hat { i } -\hat { k } \right) A-{ m }^{ 2 }$

Explanation

$\vec{DA}=-2\cos 30\hat i-2\sin 30\hat k=(-\sqrt{3}\hat i-\hat k)$

$\vec{AB}=2\hat j$

$\therefore \vec{M}=i(\vec{DA}\times\vec{AB})$

$=\dfrac{1}{2}](-\sqrt{3}\hat i-\hat k)\times(2\hat j)]$

$=-\sqrt{3}\hat k+i$

$=(i-\sqrt{3}\hat k)\ A-m^2$

Magnetic field strength at the centre of regular pentagon made of a conducting wire of uniform cross section area as shown in figure is : (i amount of current enters at A and leaves at E)

0%
$\frac{5\mu_0 i}{4\pi a} [2sin \frac{72^0}{2}]$
0%
0
0%
$\frac{3\mu_0 i}{4\pi a} [2sin \frac{72^0}{2}]$
0%
$\frac{\mu_0 i}{4\pi a} [2sin \frac{72^0}{2}]$

Explanation

$\quad In\quad this\quad the\quad magnetic\quad field\quad due\quad to\quad each\quad straight\quad conductor\quad is\\ \quad \quad \quad being\quad added\quad and\quad is\quad in\quad the\quad inward\quad direction\\ \quad \quad the\quad magnetic\quad field\quad due\quad to\quad one\quad straight\quad conductor\quad B=\quad \frac { { \mu }_{ 0 }i(sin\alpha \quad +sin\beta ) }{ 4\pi r } \\ \quad \quad Here\quad both\quad \alpha =\beta ={ 72/2 }^{ 0 }\quad degrees\\ \quad \quad \quad \quad \quad Now\quad putting\quad value\quad B=\quad \frac { { \mu }_{ 0 }i(2sin{ 72/2 }^{ 0 }) }{ 4\pi r } \\ \quad \quad \quad \quad \quad \quad Now\quad total\quad magnetic\quad field\quad at\quad the\quad centre\quad is\quad 4B=\quad 4\frac { { \mu }_{ 0 }i(2sin{ 72/2 }^{ 0 }) }{ 4\pi r } \\ \quad \quad \quad \quad \quad \quad where\quad r\quad is\quad \frac { 4a }{ 5 } \quad therefore\quad total\quad magnetic\quad field\quad becomes=\frac { 5{ \mu }_{ 0 }i(2sin{ 72/2 }^{ 0 }) }{ 4\pi a } \\ \quad therefore\quad option\quad A$

The correct Biot-Savart law in vector form is?

0%
$d\vec{B}=\dfrac{\mu_0}{4\pi}\dfrac{I(d\vec{l}\times \vec{r})}{r^2}$
0%
$d\vec{B}=\dfrac{\mu_0}{4\pi}\dfrac{I(d\vec{l}\times \vec{r})}{r^3}$
0%
$d\vec{B}=\dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}}{r^2}$
0%
$d\vec{B}=\dfrac{\mu_0}{4\pi}\cdot \dfrac{Id\vec{l}}{r^3}$

Explanation

Biot-savart law is given by

$dB=\dfrac{\mu_0}{4\pi} \dfrac{idlsin\theta}{r^2}$

vector form of biot-savart law is given by,

$dB=\large { \dfrac{\mu_0}{4\pi}}\times \dfrac{i\overrightarrow{dl}\times \overrightarrow{r}}{\overrightarrow{r^3}}$

SI unit B is tesla

magnetic field due to current carrying wire is given by,

$B\rightarrow \int\dfrac{\mu_0}{4\pi}\dfrac{i(\overrightarrow{dl} \times \overrightarrow{r})}{r^3}$

1262205_1619167_ans_a6dc26c0825e490b945f3cc401abb5d3.PNG

A long straight wire of circular cross-section is made of a non-magnetic material. The wire is of radius

$a$ .The wire carries a current

$I$ which is uniformly distributed over its cross-section. The energy stored per unit length in the magnetic field contained within the wire is

0%
$U=\dfrac {\mu_{0}I^{2}}{8 \pi}$
0%
$U=\dfrac {\mu_{0}I^{2}}{16 \pi}$
0%
$U=\dfrac {\mu_{0}I^{2}}{4 \pi}$
0%
$U=\dfrac {\mu_{0}I^{2}}{2 \pi}$

Explanation

We use a standard right handed coordinate system

$\left( {r,\phi ,z} \right)$

Since the current is entirely axial, the Boit-savart law implies that

${B_z} = 0$ every where

Symmetry consideration require that the

$r$ and

$\phi$ components of

$B$ depends upon

$r:B={ B_{ r } }\left( r \right) \hat { r } +{ B_{ \phi } }\left( r \right) \hat { \phi }$

Now apply Gauss's law for magnetic field to the coaxial cylindrical surface

$S$ , which has length

$l$ and radius

$r$

$0 = \int_S {B.da = {B_r}\left( r \right)2\pi rl}$

Therefore,

${B_r}\left( r \right) = 0$

Amper's law to coaxial circular path

$C$

$\oint_c {B.dI = 2\pi r{B_\phi }\left( r \right) = {\mu _0}i\left( r \right)}$

$........(1)$

Here

$i(r)$ is the current passing through the surface enclosed by

$C$

Since, the current density is uniform we have,

$\begin{array}{l} \frac { { i\left( r \right) } }{ i } =\frac { { \pi { r^{ 2 } } } }{ { \pi { a^{ 2 } } } } \\ i\left( r \right) =\frac { { { r^{ 2 } }i } }{ { { a^{ 2 } } } } \end{array}$

$...(2)$

Equation

$1$ and

$2nd$ implies that

${B_\phi }\left( r \right) = \frac{{{\mu _0}ir}}{{\left( {2\pi {a^2}} \right)}}$

Now integrate the magnetic energy density over the volume of the wire. The magnetic energy

$d{U_m}$ contained in the cylindrical shell of length

$1$ , inner radius

$r$ and outer radius

$r+dr$ is

$\begin{array}{l} d{ U_{ m } }={ \in _{ m } }UV=\frac { { { B_{ \phi } }^{ 2 }\left( r \right) } }{ { 2{ \mu _{ 0 } } } } 2\pi r1dr \\ =\frac { { { \mu _{ 0 } }{ i^{ 2 } }1 } }{ { 4\pi { a^{ 4 } } } } { r^{ 3 } }dr \end{array}$

Hence, the magnetic energy per unit length contained in the wire is

$\frac{{{U_m}}}{1} = \frac{1}{1}\int {d{U_m} = \frac{{{\mu _0}{i^2}}}{{4\pi {a^4}}}\int_0^a {{r^3}dr = \frac{{{\mu _0}{r^2}}}{{4\pi {a^4}}}\left( {\frac{{{r^4}}}{4}|_0^a} \right)} }$

$= \frac{{{\mu _0}{i^2}}}{{16\pi }}$

Hence, Option B is the correct answer.

1100686_1022112_ans_3f7e524760c14641b0786f36f706285c.jpg

The current carrying wire of length

$l$ and current

$I$ is coiled into a circular coil of

$n$ number of turns. The magnetic dipole moment of the coil is

0%
$\cfrac{Il^{2}}{{\pi}n}$
0%
$\cfrac{Il^{2}}{2{\pi}n}$
0%
$\cfrac{nIl^{2}}{4\pi}$
0%
$\cfrac{Il^{2}}{4{\pi}n}$

Explanation

When turned into coil, its radius becomes;

$n(2 \pi r) = l$

$\Rightarrow r = \cfrac{l}{2 \pi n}$

$\mu = niA$

$\mu = \cfrac{ni \times \pi l^2}{4 \times \pi^{2} \times n^2}$

$\mu = \cfrac{i l^2}{4 \pi n}$

973830_1075195_ans_d7c577f1c46040ff86b6424ee6d356ec.png

When a charged particle enters in a uniform magnetic field, then its kinetic energy

0%
remains constant
0%
increases
0%
decreases
0%
becomes zero

Explanation

When a charged particle enters a magnetic field

$B$ its kinetic energy remains constant as the force exerted on the particle is:

$F=q\overrightarrow { V } \times \overrightarrow { B }$
is perpendicular to

$\overrightarrow {V}$ , so the work done by

$\overrightarrow {B}=0$ . This does not cause any change in kinetic energy.

A circular coil of radius 4 cm and of 20 turns carries a current of 3 amperes. It is placed in a magnetic field of intensity of

$0.5 \ weber /m^2$ . The magnetic dipole moment of the coil is

0%
$0.15 \ ampere-m^2$
0%
$0.3 \ ampere-m^2$
0%
$0.45 \ ampere-m^2$
0%
$0.6 \ ampere-m^2$

Explanation

$\textbf{Hint:}$ Use the magnetic dipole moment formula.

$\textbf{Step 1:}$ Given values,

Radius,

$r= 4\,cm$

Number of turns,

$N=20$

Current,

$I=3 Amp$

Magnetic field intensity,

$B=0.5 T$

$\textbf{Step 2:}$ Calculating dipole moment,

$M=NiA$

$M=20 \times \dfrac{22}{7} (4 \times 10^{-2})^2\times 3$

$M=0.3\ A-m^2$

$\textbf{Option B is correct.}$

Two long thin charged rods with charge density

$\lambda$ each are placed parallel to each other at a distance d apart. The force per unit length exerted on one rod by the other will beWhere

$\left( K=\dfrac { 1 }{ 4\pi \varepsilon_ 0 }\right)$ :

0%
$\dfrac { K2\lambda }{ d }$
0%
$\dfrac { K2\lambda ^ 2 }{ d }$
0%
$\dfrac { K2\lambda }{ d^ 2 }$
0%
$\dfrac { K2\lambda ^ 2 }{ d^ 2 }$

Explanation

Given,

Charge density of each rod, $\lambda$ (charge per unit length)

$K=\dfrac{1}{4\pi {{\varepsilon }_{o}}}$

Separation between rod, $d$

Electric field (at distance d) by a uniform charged rod, $E=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{2\lambda }{d}=\dfrac{2K\lambda }{d}$

Total charge on second rod $\text{=}\ \text{charge}\ \text{density}\ \times\ \text{length}\ \text{=}\ \lambda L$

Force on charged rod placed in electric field, $F=(\lambda L)E=\dfrac{2KL{{\lambda }^{2}}}{d}$

Force per unit length due to rod is $\dfrac{F}{L}=\dfrac{2K{{\lambda }^{2}}}{d}$

The maximum energy of a deuteron coming out a cyclotron is

$20\ MeV$ . The maximum energy of proton that can be obtained from this accelerator is:

0%
$10\ MeV$
0%
$20\ MeV$
0%
$30\ MeV$
0%
$40\ MeV$

Explanation

In cyclotron output energy of particles coming out for deutron is

${{E}_{d}}=\dfrac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2{{m}_{D}}}$

${{E}_{P}}=\dfrac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2{{m}_{P}}}$

Taking ratios,

$\dfrac{{{E}_{d}}}{{{E}_{p}}}=\dfrac{{{m}_{p}}}{{{m}_{d}}}$ $\because {{m}_{p}}=1.6\times {{10}^{-27}}\,\,\And \,\,{{m}_{d}}=3.2\times {{10}^{-27}}$

$\dfrac{20}{{{E}_{P}}}=\dfrac{1.6\times {{10}^{-27}}}{3.2\times {{10}^{-27}}}$

${{E}_{p}}=40\,\,MeV$

A particle of mass m, charge

$q$ and kinetic energy

$t$ enters a transverse uniform magnetic field of induction B. After 3 s, the kinetic energy of the particle will be

0%
$3T$
0%
$2T$
0%
$T$
0%
$4T$

A magnetic moment m oscillating freely in earth's horizontal magnetic field makes a oscillation per minute. If the magnetic moment is quadrupled and the earth's field doubled, the number of oscillation made per minute would becomes

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

Explanation

No. of oscillation per min= ${1 \over {2\pi }}\sqrt {{{M{B_H}} \over I}}$

$n\alpha \sqrt {M{B_H}}$

$M \to 4times$

${B_H} = 2times$

$v \to \sqrt 8 times$

${v^1} = \sqrt 8 V$

$= 2\sqrt 2 n$

The magnetic field near a current carrying conductor is given by

0%
Coulomb's law
0%
Lenz' law
0%
Biot-Savart's law
0%
Kirchhoff's law

Explanation

The formula is

$B = \dfrac{\mu _o}{4\pi} I \dfrac{d\vec l \times \vec r}{r^3}$

It is Biot-sarvart's law

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 9 - 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 9 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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