MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 4

A north pole of strength

$\theta$ A m, is moved around a circle or radius 10 cm which lies around a long straight conductor carrying a current of 10 A. The work done in moving it by 2 revolutions is

0%
4 $\theta \mu$ J
0%
25.12 $\theta \mu$ J
0%
400 $\theta \mu$ J
0%
0.4 $\theta \mu$ J

A particle consisting of two electrons is moving in a magnetic field of

$(3i+2j) T$ with a velocity

$5 \times 10^{5}\hat{i}m/s$ . The magnetic force acting on the particle will be

0%
$3.2\times 10^{-13}\hat{k} \ dyne$
0%
$3.2\times 10^{13}\hat{k} \ N$
0%
$zero$
0%
$3.2\times 10^{-13}\hat{k} \ N$

Explanation

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

$\bar F=3\cdot 2\times 10^{-19}(5\times 10^5 \hat i\times (3 \hat i+2 \hat j))$

$=3\ . 2\times 10^{-13} \ \hat k \ N$

A magnet of strength

$\mu$ A m, is moved around a circle of radius 10 cm which lies around a long straight conductor carrying a current of 10 A. The work done is nearly

0%
4 $\mu$ J
0%
40 $\mu$ J
0%
400 $\mu$ J
0%
0.4 $\mu$ J

If the distance as well as the current in each of two parallel current carrying wires is doubled, the force per unit length acting between them becomes

0%
doubled
0%
remains same
0%
quadrupled
0%
halved

Explanation

Force per unit length

$(F)$ between two current carrying wires is

$F=\dfrac { \mu_o{ I }_{ 1 }{ I }_{ 2 } }{ 2A\pi }$

$N$
When both

$I_{1}$ ,

$I_{2}$ and

$r$ are doubled then

$\Rightarrow F'=$

$\dfrac { \mu_o(2{ I }_{ 1 })(2{ I }_{ 2 }) }{ 2A(2{ \pi }_{ 1 }) } = 2F$

Two parallel conductors carrying 5A each, repel with a force of 0.25 N

$m^{-1}$ . The distance between them is

0%
$4\times 10^{-5}m$
0%
$3\times 10^{-5}m$
0%
$2\times 10^{-5}m$
0%
$1\times 10^{-5}m$

Explanation

$\dfrac {df}{dl}=\dfrac {\mu_0 i_1i_2}{2\pi r}$

$0\cdot 25=\dfrac {2\times 10^{-7}\times 5\times 5}{r}$

$r=2\times 10^{-5}m$

Two long straight wires carry currents 5 A and 10 A in the same direction are separated by 20 cm. The force between wires is

0%
$5\times 10^{-5}N/m$ attractive
0%
$5\times 10^{-5}N/m$ repulsive
0%
$2\times 10^{-5}N/m$ attractive
0%
$2\times 10^{-5}N/m$ repulsive

Explanation

$\displaystyle F=\frac { Mo }{ 2A } \times \frac { { i }_{ 1 }{ i }_{ 2 } }{ R } =2\times 10^{ -7 }\times \frac { 5\times 10 }{ 20\times { 10 }^{ -2 } }$

$=5\times { 10 }^{ -5 }N/M$
As both currents are in same direction, force is attractive.

The magnitude of the force between a pair of conductors, each of length

$110\ cm$ , carrying a current of

$10\ A$ each and separated by a distance of

$10 cm$ is (

$\mu_{0}=4\pi\times 10^{-7}H/m$ )

0%
$55\times 10^{-5}N$
0%
$44\times 10^{-5}N$
0%
$33\times 10^{-5}N$
0%
$22\times 10^{-5}N$

Explanation

$F=\dfrac {\mu_0 i_1i_2}{2\pi r}(l)$

$=\dfrac{2\times 10^{-7}\times 10\times 10\times 1.1}{\cdot 1}$

$=22\times 10^{-5} N$

A charged particle, having charge

$q$ accelerated through a potential difference

$V$ enters a perpendicular magnetic field in which it experiences a force

$F$ . If

$V$ is increased to

$5 V$ ,the particle will experience a force

0%
$F$
0%
$5F$
0%
$0.2F$
0%
$\sqrt{5}F$

Explanation

$\dfrac {1}{2}mv^2=qV$

$v=\sqrt {\dfrac {2qV}{m}}$

$F_1=qvB=q\sqrt{ \dfrac {2qV}{m}} B$

$F_2=q\sqrt {\dfrac {2q\times 5V}{m}} B$

$F_2=\sqrt 5 F_1$

A particle of charge

$q$ and mass

$m$ moving with a velocity

$v$ along

$x-$ axis enters the region

$x > 0$ with uniform magnetic field

$B$ along the

$\hat k$ direction. The particle will penetrate in this region in the

$x-$ direction up to a distance

$d$ equal to

0%
$Zero$
0%
$\dfrac{mv}{qB}$
0%
$\dfrac{2mv}{qB}$
0%
$Infinity$

Explanation

Lorentz force

$F=q(\vec{v}\times\vec{B})=q(v\hat{i}\times B\hat{k}) =-qvB\hat{j}$
The particle move in a circle, the plane of the circle is perpendicular to the direction of magnetic field.
Centripetal force is provided by the Lorentz force.

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

$\Rightarrow r =\dfrac{mv}{qB}$

Two conductors each of length

$12m$ lie parallel to each other in air. The centre to centre distance between the two conductors is

$15\times 10^{-2}m$ and the current in each conductor is

$300A$ . The force in newton tending to pull the conductors together is:

0%
$14.4 N$
0%
$1.44 N$
0%
$144 N$
0%
$0.144 N$

Explanation

$F=\dfrac {\mu_0 i_1i_2l}{2\pi r}$

$=\dfrac {2\times 10^{-7}\times 300\times 300\times12}{15\times 10^{-2}}$

$=1\cdot 44 N$

The force between two parallel conductors, each of length

$50\ cm$ and distant

$20\ cm$ apart is

$100$ N. If the current in one conductor is double that in another one then the values will respectively be(Given

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

0%
$10^4A$ and $2\times 10^4A$
0%
$50A$ and $100A$
0%
$10A$ and $20 A$
0%
$5A$ and $10A$

Explanation

$F=\dfrac {\mu_0 i_1i_2 l}{2\pi r}$

$100=\dfrac {2\times 10^{-7}\times 2i\times 1\times \cdot 5}{\cdot 2}$

$10^8=i^2$

$i=10^4 Amp$
Current in other wire

$=2i=2\times10^4 Amp$

The force per unit length between two long straight conductors carrying currents 3 A each in the same direction and separated by a distance of 2.0 cm is :

0%
$9\times 10^{-7}N/m$
0%
$9\times 10^{-6}N/m$
0%
$9\times 10^{-5}N/m$
0%
$9\times 10^{-4}N/m$

Explanation

The force per unit length between two long straight current carrying conductors is given by

$F=\dfrac{\mu_0 I_1I_2}{2\pi d}$

Here, current in the conductors

$I_1=I_2=3 A,$

Distance between the two conductors,

$d=2 cm=0.02 m$ ,

So,

$F=\dfrac{4\pi \times 10^{-7}\times 3\times 3}{2\pi\times 0.02}=9\times 10^{-5}N/m$

Magnetic induction at the centre of a circular loop of area

$\pi m^2$ is

$0.1T$ . The magnetic moment of the loop is :
(

$\mu _{0}$ is permeability of air)

0%
$\displaystyle \frac{0.1\pi }{\mu _{0}}$
0%
$\displaystyle \frac{0.2\pi }{\mu _{0}}$
0%
$\displaystyle \frac{0.3\pi }{\mu _{0}}$
0%
$\displaystyle \frac{0.4\pi }{\mu _{0}}$

Explanation

$B$ at centre of loop =

$\dfrac { { \mu }_{ 0 }i }{ 2r }$
Given area of loop is A =

$\pi$

$\Rightarrow r=1\ m$

$0.1 \ T=$

$\dfrac { { \mu }_{ 0 }i }{ 2\times 1 }$

$i =$

$\dfrac { 0.2 }{ \mu_0 }$
Magnetic moment

$= i \times A =$

$\dfrac { 0.2 }{ \mu_0 } \times A\ =\ \dfrac { 0.2\pi }{ \mu_0 }$

A circular coil of wire of

$n$ turns has a radius

$r$ and carries a current

$i$ . Its magnetic dipole moment is

$M$ . Now the coil is unwound and again rewound into a circular coil of half the initial radius and the same current is passed through it, then the dipole moment of this new coil is :

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

Explanation

The length of remains same

$N_1\pi*r=N_2\pi*r/2$

$N_2=2n$

$M=NAI=nI\pi*r^2$

$M_2=N_2I*\pi*(r/2)^2=2n*I\pi*(r^2/4)$

$\dfrac{M_2}{M}=\dfrac{1}{2}$

$M_2=\dfrac{M}{2}$

The distance between two parallel wires carrying current of

$1A$ is

$1m$ . The force per unit length between the conductors is

0%
$2\times 10^{7}Nm^{-1}$
0%
$2\times 10^{-7}Nm^{-1}$
0%
$4\times 10^{7}Nm^{-1}$
0%
$4\times 10^{-7}Nm^{-1}$

Explanation

$F=\dfrac { \mu_0 }{ 2Ad } { i }_{ 1 }{ i }_{ 2 }$

$=\dfrac { 2\times { 10 }^{ -7 }\times 1\times 1 }{ 1 } =\quad 2\times 10^{ -7 }{ N }/{ m }$

In an electric motor, wires carrying a current of

$5A$ are placed at right angles to a magnetic field of induction

$0.8 T$ . If each wire has length of

$20cm$ , then the force acting on each wire is :

0%
$0.2N$
0%
$0.4N$
0%
$0.6N$
0%
$0.8N$

Explanation

$F= Bil = 0.8 \times 5 \times 20 \times 10^{-2} = 0.8N$

An infinite long straight wire is bent into a semicircle of radius

$R$ , as shown in the figure. A current

$I$ is sent through the conductor. The magnetic field at the centre of the semicircle is :

0%
$\infty$
0%
$0$
0%
$\dfrac{\mu _{0}}{4R}I$
0%
$\dfrac{\mu _{0}I}{4\pi R}(\pi +1)$

Explanation

Magnetic field due to straight portion of the wire

$=0$

Magnetic field due to circular portion of the wire:

$=\cfrac { 1 }{ 2 } \left( \cfrac { { \mu }_{ 0 }I }{ 2R } \right) \\ =\cfrac { { \mu }_{ 0 }I }{ 4R }$

Hence the correct answer is

$\cfrac { { \mu }_{ 0 }I }{ 4R }$

Two long parallel copper wires carry currents of

$5A$ each in the opposite direction . If the wires are separated by a distance of

$0.5 m$ , then the force between the two wires is

0%
$10^{-5}N$ attractive
0%
$10^{-5}N$ repulsive
0%
$2\times 10^{-6}N$ attractive
0%
$2\times 10^{-5}N$ repulsive.

Explanation

$F=\dfrac { \mu \ { i }_{ 1 }{ i }_{ 2 } }{ 2 \pi \ d } \quad =\quad \dfrac { 2\times { 10 }^{ -7 }\times 5\times 5 }{ 0.5 } ={ 10 }^{ -5 }N$

Since currents are opposite so the force is repulsive.

The magnetic field at the centre of circular loop in the circuit carrying current

$I$ shown in the figure is :

0%
$\dfrac{\mu _{0}}{4\pi }\dfrac{2I}{r}(1+\pi )$
0%
$\dfrac{\mu _{0}}{4\pi }\dfrac{2I}{r}(\pi-1 )$
0%
$\dfrac{\mu _{0}}{4\pi }\dfrac{2I}{r}$
0%
$\dfrac{\mu _{0}}{4\pi }\dfrac{I}{r}(\pi+1 )$

Explanation

Magnetic field due to circular loop is

${ B }_{ 1 }=\dfrac { { \mu }_{ 0 }I }{ 2r } = \dfrac { { \mu }_{ 0 }I }{ 4\pi r } 2\pi$

field due to straight wire is

${ B }_{ 2 }=\dfrac { { \mu }_{ 0 }I }{ 2\pi r } = \dfrac { { \mu }_{ 0 }I }{ 4\pi r } 2$

$\because$ net field

$={ B }_{ 1 }-{ B }_{ 2 }=\dfrac { { \mu }_{ 0 }I }{ 4\pi r } \quad (2\pi-2)$

$=\dfrac { { \mu }_{ 0 } }{ 4\pi } \times \dfrac { 2I }{ r } \quad (\pi-1)$

A current carrying small loop of one turn behaves like a small magnet. If

$A$ be its area,

$M$ its magnetic moment, the current in the loop will be

0%
$\dfrac{M}{A}$
0%
$\dfrac{A}{M}$
0%
$MA$
0%
$A^{2}M$

Explanation

Magnetic moment

$M= i \times A \implies i=M/A$

Two long straight parallel wires separated by a distance, carrying equal currents exert a force F per unit length on each other. If the distance of separation is doubled, and the current in each is halved, the force per unit length, between them will be :

0%
F
0%
F/2
0%
F/4
0%
F/8

Explanation

$F=\dfrac { \mu_o{ i }_{ 1 }{ i }_{ 2 } }{ 2\pi d }$

by halfing each current, we get and distance is doubled, then

$F=\dfrac { \mu_o\dfrac { { i }_{ 1 } }{ 2 } \quad \dfrac { { i }_{ 2 } }{ 2 } }{ 2\pi \quad (2d) } =\dfrac { 1 }{ 8 } \quad \bigg (\dfrac { \mu_o{ i }_{ 1 }{ i }_{ 2 } }{ 2\pi d } \bigg )=\dfrac { F }{ 8 }$

Two identical coils carry equal currents have a common centre and their planes are at right angles to each other. The ratio of the magnitude of the resultant magnetic field at the centre and the field due to one coil is :

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

Explanation

The magnetic field due to each coil is equal in magnitude and perpendicular to each other

Let magnetic field due to one of them be B

Then the resultant magnetic field due to 2 coil =

$\sqrt{ B^{2} +B^{2}} = \sqrt{2} B$

The ratio of the magnitude of the resultant magnetic field at the centre and the field due to one coil =

$\sqrt{2} B : B = \sqrt{2} :1$

Find the magnetic field due to conducting wire at point

$O,$ at centre of semicircle of radius

$r$ and carrying a current

$i$ as shown in the figure.

0%
$\dfrac{\mu _{0}i}{4r}$
0%
$\dfrac{\mu _{0}i}{4r}(1+2\pi )$
0%
$\dfrac{\mu _{0}i}{4r}(\pi-2 )$
0%
$\dfrac{\mu _{0}i}{4\pi r}(\pi+2 )$

Explanation

$\left | B \right |$ at centre =

$\left | B \right |$ due to semicircle wire+

$2\times$

$\left | B \right |$ due semi infinite wire

$=\dfrac{\mu _{0}i}{4r}+\dfrac{2\mu _{0}i}{4\pi r}$

$=\dfrac{\mu _{0}i}{4r}\times \left ( 1+\dfrac{2}{\pi } \right )$

In the given figure the magnetic induction at the point O is

0%
$\displaystyle \frac{\mu _{0}I}{4\pi r}$
0%
$\displaystyle \frac{\mu _{0}I}{4r}+\frac{\mu _{0}I}{2\pi r}$
0%
$\displaystyle \frac{\mu _{0}I}{4r}+\frac{\mu _{0}I}{4\pi r}$
0%
$\displaystyle \frac{\mu _{0}I}{4r}-\frac{\mu _{0}I}{4\pi r}$

Explanation

Magnetic field due to circular arc on its center

$\dfrac{\mu _{0}I}{4\pi r}\times \theta$

& due to infinite current carrying straight wire at a distance r from it=

$\dfrac{\mu _{0}}{2\pi r}$

$\left | B \right | = \dfrac{\mu _{0}I}{4\pi r}\times \pi + \dfrac{\mu _{0}}{4\pi r}$

$\dfrac{\mu _{0}I}{4 r} + \dfrac{\mu _{0}}{4\pi r}$

The magnetic field at the center of the coil of radius

$r$ and carrying a current

$I$ as shown in the figure is : (the wires crossing at

$P$ are insulated from each other)

0%
$\dfrac{\mu _{0}}{4\pi}\dfrac{2I}{r}(1+\pi )$
0%
$\dfrac{\mu _{0}}{4\pi}\dfrac{2I}{r}(\pi-1 )$
0%
$\dfrac{\mu _{0}}{4\pi}\dfrac{2I}{r}(\pi^{2}+1 )$
0%
$\dfrac{\mu _{0}}{4\pi}\dfrac{2\pi I}{r}$

Explanation

Given, radius of coil

$= r$ , current flowing

$= I$

Formula used,

i) Magnetic field due to infinite length wire,

$B_{01}=\dfrac { { \mu }_{ 0 }I }{ 2\pi r }$

ii) Magnetic field due to circular loop,

$B_{02}=\dfrac { { \mu }_{ 0 }I }{ 2r }$

$\therefore$ Total magnetic field at 0 is,

$B=B_{01}+B_{02}$

$=\dfrac { { \mu }_{ 0 }I }{ 2\pi r } +\dfrac { { \mu }_{ 0 }I }{ 2r }$

$=\dfrac { { \mu }_{ 0 }I }{ 2r } \left( \dfrac { 1+\pi }{ \pi } \right)$

$B=\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { 2I }{ r } \left( 1+\pi \right)$

Two identical circular loops each of radius

$r$ and carrying a current

$i$ are arranged concentric with each other and in perpendicular planes as shown in the given figure. The magnetic field

$\vec{B}$ at the common center is :

0%
$\dfrac{\mu _{0}i}{2r}$
0%
$\dfrac{\mu _{0}i}{\sqrt{2}r}$
0%
$zero$
0%
$\dfrac{\mu _{0}i}{r}$

Explanation

Given,

i) two coils (circular loop) arranged in perpendicular planes.

ii) The radius of each loop

$=r$

iii) The magnetic field at common centre =?

We know that,

Induced magnetic field due to current flowing in a plane are mutually perpendicular. Thus,

$B_1$ is perpendicular to loop 1 plane

$B_2$ is perpendicular to loop 2 plane

Resultant at centre,

$B_r=\sqrt { { B }_{ 1 }^{ 2 }+{ B }_{ 2 }^{ 2 } }$

and

$B_1=\dfrac { { \mu }_{ 0 }I }{ 2r } ={ B }_{ 2 }$

$\therefore B_r=\sqrt { 2 } { B }_{ 1 }$

$=\sqrt { 2 } \dfrac { { \mu }_{ 0 }I }{ 2r }$

$B_r=\dfrac { { \mu }_{ 0 }I }{ \sqrt { 2r } }$

The magnetic field at the centre of a circular coil of radius r is

$\pi$ times that due to a long straight wire at a distance r from it, for equal currents. Figure shows three cases. In all cases the circular path has radius r and straight ones are infinitely long. For same current the B field at the centre P in cases 1,2,3 has the ratio :

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

Explanation

Magnetic field in case 1 is

${ B }_1= \dfrac { { \mu }_{ 0 } I}{ 2 r } \left( \dfrac { \pi }{ 2 \times 2 \pi} +\dfrac{1}{2\pi} \right)$ outside the plane of paper

Magnetic field in case 2 is

${ B }_2= \dfrac { { \mu }_{ 0 } I}{ 4\pi r }$ inside the plane of paper

Magnetic field in case 3 is

${ B }_3= \dfrac { { \mu }_{ 0 } I}{ 2r } \left( \dfrac { 3\pi }{ 2 \times 2 \pi} -\dfrac{1}{2\pi} \right)$ inside the plane of paper

$\therefore { B }_{ 1 }:{ B }_{ 2 }:{ B }_{ 3 }$ =

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

The wire shown in figure carries a current of

$40A$ . If

$r=3.14cm$ the magnetic field at point p will be :

0%
$1.6\times 10^{-3}T$
0%
$3.2\times 10^{-3}T$
0%
$6\times 10^{-4}T$
0%
$4.8\times 10^{-3}T$

Explanation

Magnetic field due to circular arc is given by,

$\left | B \right |=\dfrac{\mu _{0}I }{4\pi r}\theta$ =

$\dfrac{\mu _{0}I\times 3\pi }{4\pi r \times 2}$ =

$\dfrac{3 \mu _{0}I}{8r}$
=

$\dfrac{3\times 4\pi \times 10^{-7}\times 40}{8\times 3.14\times 10^{-2}}$
=

$6\times 10^{-4}T$

A charged particle, moving at right angle to a uniform magnetic field, starts moving along a circular arc of radius of curvature

$r$ . In the field it now penetrates a layer of lead and loses

${\dfrac{3}{4} }^{th}$ of it's initial kinetic energy. The radius of curvature of it's path now will be :

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

Explanation

Initially,

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

$r=\dfrac{\sqrt{2K_{e}m}}{qB}$

Now

${K_{e}}'=\dfrac{1}{4}K_{e}$

${r}'=\dfrac{\sqrt{2\times \bigg (\dfrac{1}{4}\bigg )K_{e}m}}{qB}$

=

$\dfrac{r}{2}$

Two particles

$X$ and

$Y$ having equal charges, after being accelerated through the same potential differences, enter in a region of uniform magnetic field and describe circular paths of radii

$R_{1}$ and

$R_{2}$ respectively. The ratio of the mass of

$X$ to that of

$Y$ is :

0%
$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{1/2}$
0%
$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{-1}$
0%
$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{2}$
0%
$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )$

Explanation

Since the particle is accelerated through

$V$ volts

$\because \dfrac{mv^{2}}{2}=qV$

$mv=\sqrt{2mqV}$

$R=\dfrac{mv}{qB}$
So,

$R_1=\dfrac{\sqrt{2m_{1}qV}}{qB}$

$R_2=\dfrac{\sqrt{2m_{2}qV}}{qB}$

$\dfrac{m_1}{m_2}=\bigg (\dfrac{R_1}{R_2}\bigg )^{2}$

In the given loop the magnetic field at the centre O is :

0%
$\dfrac{\mu _{0}I}{4}\left ( \dfrac{r_{1}+r_{2}}{r_{1}r_{2}} \right )$ out of the page
0%
$\dfrac{\mu _{0}I}{4}\left ( \dfrac{r_{1}+r_{2}}{r_{1}r_{2}} \right )$ into the page
0%
$\dfrac{\mu _{0}I}{4}\left ( \dfrac{r_{1}-r_{2}}{r_{1}r_{2}} \right )$ out of the page
0%
$\dfrac{\mu _{0}I}{4}\left ( \dfrac{r_{1}-r_{2}}{r_{1}r_{2}} \right )$ into the page

Explanation

$\left | B_{1} \right |=\dfrac{\mu _{0}I}{4r_{1}}$ into the plane

$\left | B_{2} \right |=\dfrac{\mu _{0}I}{4r_{2}}$ into the plane

Net

$\left | B_{1} \right |=\dfrac{\mu _{0}I}{4}\times \left (\dfrac{r_{1}+r_{2}}{r_{1}r_{2}} \right )$ into the plane

A deutron of kinetic energy

$50 keV$ is describing a circular orbit of radius

$0.5 m$ in a plane perpendicular to magnetic field

$\vec{B}$ . The kinetic energy of the proton that describes a circular orbit of radius

$0.5 m$ in the same plane with the same

$\vec{B}$ is :

0%
$25 keV$
0%
$50 keV$
0%
$200 keV$
0%
$100 keV$

Explanation

Deutron is

$_{1}H^{2}$

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

$r=\dfrac{\sqrt{2K.m}}{qB}$

$\because r_{1}=r_{2}$

$\dfrac{\sqrt{2\times 50\times 2}}{B}=\dfrac{\sqrt{2\times K.1}}{B}$

$K=100keV$

A proton of energy

$E$ is moving along a circular path in a uniform magnetic field. If an alpha particle describes the same circular path, its energy should be :

0%
$4E$
0%
$2E$
0%
$E$
0%
$0.5E$

Explanation

proton

$\dfrac{mv^{2}}{2}=E$

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

$\dfrac{\sqrt{2Em}}{qB}$
for

$\alpha$ particle,

$M=4m \ and \ Q=2q$

$r=\dfrac{\sqrt{2{E}'4m}}{2qB}$

$r=\dfrac{\sqrt{2{E}'m}}{qB}$

${E}'=E$

Electrons accelerated by a potential difference

$V$ enter a uniform magnetic field of flux density

$B$ at right angles to the field. They describe a circular path of radius

$r$ . If now

$V$ is doubled and

$B$ is also doubled, the radius of the new circular path is :

0%
$4r$
0%
$2r$
0%
$2\sqrt{2r}$
0%
$\dfrac{r}{\sqrt{2}}$

Explanation

Radius of charged particle in magnetic field is,

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

$\dfrac{mv^{2}}{2}=qV$

$mv=\sqrt{2qVm}$

$r=\dfrac{mv}{qB}=\dfrac{\sqrt{2qVm}}{qB}=\dfrac{1}{B}\times\bigg (\sqrt{\dfrac{2Vm}{q}}\bigg )$

${r}'=\dfrac{1}{2B}\times\bigg (\sqrt{\dfrac{2\times 2Vm}{q}}\bigg )=\dfrac{r}{\sqrt{2}}$

A proton moving with a velocity of

$(6\hat{i} + 8\hat{j}) \times 10^{5} \ ms^{-1}$ enters uniform magnetic field of induction

$5 \times 10^{-3} \hat{k}T$ . The magnitude of the force acting on the proton is :
(

$\hat{i}, \hat{j}$ and

$\hat{k}$ are unit vectors forming a right handed triad)

0%
$0$
0%
$8\times 10^{-16}N$
0%
$3\times 10^{-16}N$
0%
$4\times 10^{-16}N$

Explanation

Force on a charged particle in magnetic field is given by,

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

$1.6\times 10^{-19}((6\hat{i}+8\hat{j})\times 5\hat{k})\times10^{5}\times 10^{-3}$

$1.6\times 10^{-19}\times 10^{-2}(-30\hat{j}+40\hat{i})$

$1.6\times 10^{-16}(-3\hat{j}+4\hat{i})$

${\left | F \right |}=1.6\times 10^{-16}\sqrt{3^{2}+4^{2}}$

$1.6\times 10^{-16}\times 5 N$

$8\times 10^{-16} N$

A proton, a deuteron and an

$\alpha$ particle are accelerated through same potential difference and then they enter in a normal uniform magnetic field, the ratio of their kinetic energies will be:

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$2:1:3$
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$1:1:2$
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$1:1:1$
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$1:2:4$

Explanation

Kinetic energy obtained

$=qV$ where

$q$ is the charge and

$V$ is the potential through which the particle is accelerated.
Also, the force due to the magnetic field is perpendicular to the direction of motion. So it cant change the speed and consequentially Kinetic energy of the particles will remain same after they enter a normal magnetic field.
In the above problem,

$V$ is constant.
So KE

$\propto q$ .
The charges of the proton, deuteron and the alpha particle are in the ratio

$1 : 1 : 2$
So, their kinetic energies are in the ratio

$1 : 1 : 2$

If a particle of charge

$10^{-12}C$ moving along the x-axis with a velocity

$10^{5}$ m/s. experiences a force of

$10^{-10}N$ in y-direction due to magnetic field, then the minimum magnetic field is.

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$6.25\times 10^{3}T$ in Z direction
0%
$10^{-15}T$ in Z direction
0%
$6.25\times 10^{-3}T$ in Z direction
0%
$10^{-3}T$ in -ve Z direction

Explanation

$f=qVB\sin \theta$

B will be minimum when

$\theta =90^{\circ}$

$B=\dfrac{f}{qV}$
=

$\dfrac{10^{-10}}{10^{-12}\times 10^{5}}$
=

$10^{-3}T$ in Z-direction

Acceleration experienced by a particle with specific charge

$1 \times10^{7} C/kg$ when fired perpendicular to a magnetic field of induction

$100 T$ with a velocity

$10^{5} ms^{-1}$ is :

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$10^{8} ms^{-2}$
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$10^{-6} ms^{-2}$
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$10^{14} ms^{-2}$
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$10^{-8} ms^{-2}$

Explanation

Specific charge

$\left(\dfrac{q}{m}\right)=10^{7} C/kg$

$\left | B \right |=100T$

$f=qVB \sin \theta$

$q=\dfrac{f}{m}$

=

$\dfrac{qVB \sin \theta}{m}$

=

$\dfrac{10^{7}\times 10^{5}\times 100 \times \sin 90^{\circ}}{1}$

=

$10^{14} ms^{-2}$

A proton of energy

$8eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

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$4eV$
0%
$2eV$
0%
$8eV$
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$6eV$

Explanation

From the formula mentioned above, momentum of particle moving in a magnetic field

$mv=p=qBr$
Therefore, Kinetic Energy of that particle can be written as

$KE=\dfrac{p^2}{2m}=\dfrac{q^2B^2r^2}{2m}$
In the same magnetic field for the same path,

$KE \propto \dfrac{q^2}{m}$
This ratio is same for the alpha particle and the proton. (

$\dfrac{(2e)^2}{4 amu}=\dfrac{4e^2}{4 amu}=\dfrac{e^2}{amu}$ ; Here

$amu$ is the atomic mass unit)
So, in such conditions, both will have the same energy. Hence, energy of the alpha particle will be

$8 eV$ too.

An electron having energy

$10 eV$ is circulating in a path of radius

$0.1m$ having a magnetic field of

$10^{-4}T$ . The speed of the electron will be :

0%
$2.0\times (10^{6})ms^{-1}$
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$4.8\times (10^{6})ms^{-1}$
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$2.0\times (10^{12})ms^{-1}$
0%
$4.8\times (10^{12})ms^{-1}$

Explanation

Magnetic force provides necessary centripetal force,

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

$qVB=\dfrac{2KE}{r}$

$V=\dfrac{2KE}{qBr}$

=

$\dfrac{2\times 10}{10^{-4}\times 0.1}$

=

$2.0 \times10^{6}ms^{-1}$

An electron moves with a speed

$2\times 10^{5}\ m/s$ along the positive x-direction in the presence of a magnetic induction

$\vec{B}=\hat{i}+4\hat{j}-3\hat{k}\ T$ . The magnitude of the force experienced by the electron in newton is :
(charge on the electron

$=1.6 \times10^{-19}C$ )

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$1.18\times 10^{-13}$
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$1.28\times 10^{-13}$
0%
$1.6\times 10^{-13}$
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$1.72\times 10^{-13}$

Explanation

Force on charge particle in magnetic field is given by,

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

$V=2\times 10^5 \hat{i}$

$1.6\times 10^{-19}(V\hat{i}\times (\hat{i}+4\hat{j}-3\hat{k}))$

$1.6\times 10^{-19}(4V\hat{k}+3V\hat{j})$

$1.6\times 10^{-19}(4\hat{k}+3\hat{j})\times 2\times 10^{5}$

${ |F| }=1.6\times 10^{-13}N$

Two particles having same charge and

$KE$ enter at right angles into the same magnetic field and travel in circular paths of radii

$2 cm$ and

$3 cm$ respectively. The ratio of their masses is :

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$2 : 3$
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$3 : 2$
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$4 : 9$
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$9 : 4$

Explanation

Radius of charged particle in magnetic field is,

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

$\dfrac{mv^{2}}{2}=K_{e}$

$r=\dfrac{\sqrt{2K_{e}m}}{qB}$

$r_{1}=\dfrac{\sqrt{2K_{e}m_{1}}}{qB}$

$r_{2}=\dfrac{\sqrt{2K_{e}m_{2}}}{qB}$

$\dfrac{4}{9}=\dfrac{m_{1}}{m_{2}}$

A proton of energy

$2 MeV$ is moving perpendicular to uniform magnetic field of

$2.5 T$ . The force on the proton is : (

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

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

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$2.5\times 10^{-16}N$
0%
$8\times 10^{-11}N$
0%
$2.5\times 10^{-11}N$
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$8\times 10^{-12}N$

Explanation

$F=qvB$

$\dfrac{mv^{2}}{2}=K_{e}$

$v=\sqrt{\dfrac{2K_{e}}{m}}$

$\therefore f=q\times \sqrt{\dfrac{2K_{e}}{m}}\times B$
=

$1.6\times 10^{-19}\times \sqrt{\dfrac{2\times 2\times 10^{6}\times 1.6\times 10^{-19}}{1.6\times 10^{-27}}}\times 2.5$
=

$8\times 10^{-12} N$

A horizontal wire carries 200 amp current below which another wire of linear density

$20\times 10^{-3}kgm^{-1}$ carrying a current is kept at 2 cm distance. If the wire kept below hangs in air. The current in this wire is :

0%
100A
0%
9.8 A
0%
98 A
0%
48A

Explanation

Since wire hangs in the air, So

$mg=Bil$

$\dfrac{m}{l}=\dfrac{Bi}{g}$

$20\times 10^{-3}=\bigg (\dfrac{\mu _{0}i}{g2\pi r}\bigg )i$

$i=98A$

An electron travelling with a velocity

$\bar{V}=10^{7}i\ m/s$ enter a magnetic field of induction

$\bar{B}=\overline{2j}$ . The force on electron is

0%
$1.6\times 10^{-12}\bar{j}N$
0%
$3.2\times 10^{-12}\bar{k}N$
0%
$6.4\times 10^{-12}\bar{k}N$
0%
$-3.2\times 10^{-12}\bar{k}N$

Explanation

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

$q\times 10^{7}\times 2\times \bar{k}$
=

$1.6\times 10^{-19}\times 2\times 10^{7} \bar{k} N$
=

$3.2\times 10^{-12}\bar{k} N$

Two long parallel conductors are placed at right angles to a metre scale at the

$2 cm$ and

$4 cm$ marks, as shown in the figure. They carry currents of

$1 A$ and

$3 A$ respectively. They will produce zero magnetic field at the

(ignore the Earth's magnetic field)

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$0.5 cm$ mark
0%
$2.5 cm$ mark
0%
$1 cm$ mark
0%
$8 cm$ mark

Explanation

$|B|_{net}=0$

$(\dfrac{\mu _{0}i_{1}}{2\pi r_{1}})-(\dfrac{\mu _{0}i_{2}}{2\pi r_{2}})=0$

where

$r1$ and

$r_2$ are distances from wires respectively. Assuming the distance from wire 1 is

$x$ , then distance from wire 2 will be

$2-x$

$\dfrac{i_{1}}{x}=\dfrac{i_2}{2-x}$

$\dfrac{1}{x}=\dfrac{3}{2-x}$

$4x=2$

$x=0.5 cm$ from

$1A$ wire.
distance from the y-axis

$2+0.5=2.5cm$ .

Two infinitely long straight conductors which are held parallel to each other at a distance

$0.4m$ carry currents

$20A$ and

$10A$ in the same direction. The magnetic induction at a point midway between them is

0%
$10^{-5}T$
0%
$3\times 10^{-5}T$
0%
$2\times 10^{-5}T$
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$10^{-6}T$

Explanation

$\left | B \right |=(\dfrac{\mu _{0}i_{1}}{2\pi r_{1}})-(\dfrac{\mu _{0}i_{2}}{2\pi r_{2}})$
=

$(\dfrac{2\times 10^{-7}\times 20}{0.2})-(\dfrac{2\times 10^{-7}\times 10}{0.2})$
=

$50\times 2\times 10^{-7}T$ =

$10^{-5}T$

An experimenters diary reads as follows : A charged particle is projected in a magnetic field

$(7\hat{i}-3\hat{j}) \times10^{-3}$ T . The acceleration of the particle is found to be

$(\hat{i}+7\hat{j}) \times10^{-6}ms^{-2}$ then coefficient of

$\hat{i}$ is

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$3.0$
0%
$7.0$
0%
$5.0$
0%
$4.0$

Explanation

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

$m\bar{a}=q(\bar{V}\times \bar{B})$

$\therefore \bar{a}.\bar{B}=0$

$(\hat{i}+7\hat{j}).(7\hat{i}-3\hat{j})=0$

$7x -21=0$

$x=3$

When two infinitely long parallel wires separated by a distance of

$1m$ , each carry a current of

$3A$ , the force in newton/metre length experienced by each will be, (given

$\mu _{0}=4\pi \times 10^{-7}$ S.I. Units).

0%
$2\times 10^{-7}$
0%
$3\times 10^{-7}$
0%
$6\times 10^{-7}$
0%
$18\times 10^{-7}$

Explanation

$F = I LB$

$B=\dfrac{\mu_{o} I_{2}} {2\pi d}$

$L = 1 m$

$I = 3 A$

$I_{2} =3 A$

$d = 1 m$

$\implies F =$

$\dfrac{3\times 1\times 4\pi \times10^{-7} \times 3}{2\pi}$ =

$18\times 10^{-7}$ N

Answer. D

The ratio of magnetic field at the centre of a current carrying coil to its magnetic moment is

$x$ . If the current and radius both are doubled, the new ratio will become:

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$2x$
0%
$4x$
0%
$x/4$
0%
$x/8$

Explanation

Magnetic field at center of current carrying coil is,

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

Magnetic moment is given by:

$M=i\pi r^2$

Given ratio is:

$x=\dfrac{B}{M}$

$x=\dfrac{\mu_o}{2\pi r^3}$

On doubling the current and radius,

$x'=\dfrac{\mu_0}{2\pi \times (2r)^3}$

$\therefore x'=\dfrac{x}{8}$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 4 - 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 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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