MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 13

Two particles, each of mass

$m$ and charge

$q$ , are attached to the two ends of a light rigid rod of length

$2R$ . The rod is rotated at constant angular speed about a perpendicular axis passing through its center. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the center of the rod is

0%
$\dfrac{q}{2m}$
0%
$\dfrac{q}{m}$
0%
$\dfrac{2q}{m}$
0%
$\dfrac{q}{\pi m}$

Explanation

Current,

$i=$ (frequency) (charge)

$=\left (\dfrac {\omega}{2\pi}\right )(2q)=\dfrac {q\omega}{\pi}$

Magnetic moment,

$M=(I)(A)=\left (\dfrac {q\omega}{\pi}\right )(\pi R^2)=(q\omega R^2)$

Angular momentum,

$L=2I\omega =21(mR^2)\omega$

$\dfrac {M}{L}=\dfrac {q\omega R^2}{2(mR^2)\omega}=\dfrac {q}{2m}$

Find the magnetic moment vector of the loop

0%
$(0.1\hat i+0.05\hat j-0.05\hat k)Am^2$
0%
$(0.1\hat i+0.05\hat j+0.05\hat k)Am^2$
0%
$(0.1\hat i-0.05\hat j+0.05\hat k)Am^2$
0%
$(0.1\hat i-0.05\hat j-0.05\hat k)Am^2$

Explanation

The loop can be decomposed into sum of three loops ABEDA, BCEB and CDEC.
Magnetic moment of a loop is

$\vec{\mu }= i\vec{A}$
Area of ABEDA is

$A=a^2$

Area of each of the loop BCEB and CDEC is

$A=\dfrac{1}{2}a^2$ where a is the length of a side of the cube.

$\vec{\mu}_{ABEDA}= i\times a^2 \hat{i}$

$\vec{\mu}_{BCEB}= i\times \dfrac{1}{2}a^2 \hat{j}$

$\vec{\mu}_{CDEC}= i\times \dfrac{1}{2}a^2 \hat{k}$

$\vec{\mu}= \vec{\mu}_{ABEDA} + \vec{\mu}_{BCEB} + \vec{\mu}_{CDEC} = i\times a^2 \hat{i} + i\times \dfrac{1}{2}a^2 \hat{j} + i\times \dfrac{1}{2}a^2 \hat{k}$

$\therefore \vec{\mu}= 10 \times 10^{-2}( \hat{i} + 0.5 \hat{j} + 0.5 \hat{k})=(0.1\hat{i} + 0.05 \hat{j} + 0.05 \hat{k})\:Amp\: m^2$

Ratio of the currents

${I}_{1}$ and

${I}_{2}$ flowing through the circular and straight parts is

0%
$\cfrac { \sqrt { 3 } }{ 2\pi }$
0%
$\cfrac { 2\sqrt { 3 } }{ \pi }$
0%
$\cfrac {3 \sqrt { 3 } }{ 2\pi }$
0%
$\cfrac { 3\sqrt { 3 } }{ 2\sqrt { 2 } \pi }$

Explanation

${I}_{1}$ and

${I}_{2}$ be the currents in circular and straight parts, respectively, and

${B}_{1}$ and

${B}_{2}$ the magnetic fiels due to them, then

${ B }_{ 1 }=\cfrac { { \mu }_{ 0 }{ I }_{ 1 } }{ 2R } \times \cfrac { 2 }{ 3 } =\cfrac { { \mu }_{ 0 }{ I }_{ 1 } }{ 3R }$

${ B }_{ 2 }=\cfrac { { \mu }_{ 0 }{ I }_{ 2 } }{ 4\pi \left[ R\cos { { 60 }^{ o } } \right] } \left[ 2\sin { { 60 }^{ o } } \right] =\cfrac { \sqrt { 3 } { \mu }_{ 0 }{ I }_{ 2 } }{ 2\pi R }$

For the total field at

$'O'$ to be zero,

$\cfrac { { \mu }_{ 0 }{ I }_{ 1 } }{ 3R } =\cfrac { \sqrt { 3 } { \mu }_{ 0 }{ I }_{ 2 } }{ 2\pi R } \quad \Rightarrow \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =\cfrac { 3\sqrt { 3 } }{ 2\pi }$

From

$R=\cfrac { \rho l\rho }{ A }$

$\cfrac { { R }_{ 1 } }{ { R }_{ 2 } } =\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \cfrac { { A }_{ 1 } }{ { A }_{ 2 } } \quad \Rightarrow \quad \cfrac { V/{ I }_{ 1 } }{ V/{ I }_{ 2 } } =\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \cfrac { (\pi /4){ { d }_{ 2 } }^{ 2 } }{ (\pi /4){ { d }_{ 1 } }^{ 2 } } \quad \Rightarrow \cfrac { { d }_{ 1 } }{ { d }_{ 2 } } =\sqrt { \cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } }$

Where

${ l }_{ 1 }=2\pi R\left( \cfrac { 2 }{ 3 } \right) =\cfrac { 4 }{ 3 } \pi R;\quad { l }_{ 2 }=2R\sin { { 60 }^{ o } } =\sqrt { 3 } R$

$\Rightarrow \cfrac { { d }_{ 1 } }{ { d }_{ 2 } } =\sqrt { \cfrac { 4\pi R }{ 3\sqrt { 3 } R } \cfrac { \sqrt { 3 } R }{ 2\pi } } =\sqrt { 2 }$

The magnetic field at the origin due to the current flowing in the wire is

0%
$-\cfrac { { \mu }_{ 0 }I }{ 8\pi a } \left( \hat { i } +\hat { k } \right)$
0%
$\cfrac { { \mu }_{ 0 }I }{ 2\pi a } \left( \hat { i } +\hat { k } \right)$
0%
$\cfrac { { \mu }_{ 0 }I }{ 8\pi a } \left(- \hat { i } +\hat { k } \right)$
0%
$\cfrac { { \mu }_{ 0 }I }{ 4\pi a \sqrt{2}} \left( \hat { i } -\hat { k } \right)$

Explanation

${B}_{OD}=0$

${B}_{OB}=0$

$\quad { B }_{ AB }=\cfrac { { \mu }_{ 0 } }{ 4\pi a\sqrt { 2 } } \left[ \cos { { 45 }^{ o } } (-\hat { i } )+\cos { { 45 }^{ o } } \hat { k } \right]$

$=\cfrac { { \mu }_{ 0 } }{ 8\pi a } (-\hat { i } +\hat { k } )$

The ratio of the diameters of wires of circular and straight parts is

0%
$\cfrac{1}{\sqrt 2}$
0%
$\cfrac { 2\sqrt { 3 } }{ \pi }$
0%
$\cfrac {3\sqrt { 3 } }{ 2\pi }$
0%
$\sqrt {2}$

Explanation

${I}_{1}$ and

${I}_{2}$ be the currents in circular and straight
parts, respectively, and

${B}_{1}$ and

${B}_{2}$ the magnetic fields due to them, then

${ B }_{ 1 }=\cfrac { { \mu }_{ 0 }{ I }_{ 1 } }{ 2R } \times \cfrac { 2 }{ 3 } =\cfrac { { \mu }_{ 0 }{ I }_{ 1 } }{ 3R }$

${B }_{ 2 }=\cfrac { { \mu }_{ 0 }{ I }_{ 2 } }{ 4\pi \left[ R\cos { { 60 }^{ o } } \right] } \left[ 2\sin { { 60 }^{ o } } \right] =\cfrac {\sqrt { 3 } { \mu }_{ 0 }{ I }_{ 2 } }{ 2\pi R }$

For the total field at

$'O'$ to be zero,

$\cfrac{ { \mu }_{ 0 }{ I }_{ 1 } }{ 3R } =\cfrac { \sqrt { 3 } { \mu }_{ 0 }{ I }_{ 2 } }{ 2\pi R } \quad \Rightarrow \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =\cfrac { 3\sqrt { 3 } }{ 2\pi }$

From

$R=\cfrac { \rho l\rho }{ A }$

$\cfrac{ { R }_{ 1 } }{ { R }_{ 2 } } =\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \cfrac { { A }_{ 1 } }{ { A }_{ 2 } } \quad \Rightarrow \quad \cfrac { V/{ I }_{ 1 } }{ V/{ I }_{ 2 } } =\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \cfrac { (\pi /4){ { d }_{ 2 } }^{ 2 } }{ (\pi /4){ { d }_{ 1 } }^{ 2 } } \quad \Rightarrow \cfrac { { d }_{ 1 } }{ { d }_{ 2 } } =\sqrt { \cfrac{ { l }_{ 1 } }{ { l }_{ 2 } } \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } }$

Where

${ l }_{ 1 }=2\pi R\left( \cfrac { 2 }{ 3 } \right) =\cfrac { 4 }{ 3 } \pi R;\quad { l }_{ 2 }=2R\sin { { 60 }^{ o } } =\sqrt { 3 } R$

$\Rightarrow \cfrac { { d }_{ 1 } }{ { d }_{ 2 } } =\sqrt { \cfrac { 4\pi R }{ 3\sqrt { 3 } R } \cfrac { \sqrt { 3 } R }{ 2\pi } } =\sqrt { 2 }$

A magnetic field

$\vec B=B_0\hat j$ exists in the region

$a < x < 2a$ and

$\vec B=-B_0\hat j$ , in the region 2a < x < 3a, where

$V_0$ is a positive constant. A positive point charge moving with a velocity

$\vec v=v_0\hat i$ , where

$v_0$ is a positive constant, enters the magnetic field at

$x=a$ . The trajectory of the charge in this region can be like

Explanation

Force on a charged particle

$\vec{F} = q( \vec{v} \times \vec{B})$
From a to 2a:

$\vec{B} = B_0\hat{j}$ ,

$\vec{v}=v_0\hat{i}$

$\therefore \vec{F} = qv_0B_0(\hat{i} \times \hat{j})= qv_0B_0\hat{k}$
Thus the particle gets deflected towards z-axis.
From 2a to 3a:

$\vec{B} = -B_0\hat{j}$ ,

$\vec{v}=v_0\hat{i}$

$\therefore \vec{F} = -qv_0B_0(\hat{i} \times \hat{j})= -qv_0B_0\hat{k}$
Thus the particle gets deflected towards -ve z-axis.
Since the particle enters at

$x=a$ , the particle will have z-coordinate as zero.

$H^+, He^+$ , and

$O^{2+}$ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of

$H^+, He^+$ and

$O^{2+}$ are 1, 4, and 16 amu, respectively. Then,

0%
$H^+$ will be deflected most
0%
$O^{2+}$ will be deflected most
0%
$He^+$ and $O^{2+}$ will be deflected equally
0%
All will be deflected equally

Explanation

When the charged particles enter a magnetic field, then a force acts on the particle which will acts as centripetal force.

$qvB=\dfrac {mv^2}{r}\Rightarrow r=\dfrac {mv}{qB}$ (i)

Now,

$E=\dfrac {1}{2}mv^2\Rightarrow v=\dfrac {\sqrt {2E}}{m}$ (ii)

$\therefore r=\dfrac {m}{qB}\sqrt {\dfrac {2E}{m}}=\dfrac {\sqrt {2Em}}{qB}=\dfrac {\sqrt {2E}}{B}\times \dfrac {\sqrt m}{q}$

$\Rightarrow r\propto \dfrac {\sqrt m}{q}$

$\therefore r_{H^+}\propto \dfrac {\sqrt 1}{1}; pr_{He^+}\propto \dfrac {\sqrt 4}{1}; r_{O^+}\propto \dfrac {\sqrt {16\cdot }}{2}$

$\Rightarrow r_{H^+}\propto 1; r_{He^+}\propto 2; r_{O^+}\propto 2$

$\Rightarrow He^+$ and

$O^+$ will be deflected equally.

$H^+$ will be deflected the most since its radius is smallest.

ABCD is a square loop made of a uniform conducting wire. The current enters the loop at A and leaves at B. The magnetic field is ..........

0%
Zero only at the centre of the loop
0%
Maximum at the centre of the loop
0%
Zero at all points outside the loop
0%
Zero at all points inside the loop

A coil of radius R carries a current I. Another concentric coil of radius

$\displaystyle r\left( r<<R \right)$ caries current

$\displaystyle \frac { I }{ 2 }$ . Initially planes of the two coils are mutually perpendicular and both the coils are free to rotate about common diameter. They are released from rest from this position. The masses of the coils are M and m respectively

$\displaystyle \left( m<M \right)$ . During the subsequent motion let

$\displaystyle { K }_{ 1 }$ and

$\displaystyle { K }_{ 2 }$ be the maximum kinetic energies of the two coils respectively and let U be the magnitude of maximum potential energy of magnetic interaction of the system of the coils. Choose the correct options.

0%
$\displaystyle \frac { { K }_{ 1 } }{ { K }_{ 2 } } =\frac { M }{ m } { \left( \frac { R }{ r } \right) }^{ 2 }$
0%
$\displaystyle { K }_{ 1 }=\frac { { Umr }^{ 2 } }{ { mr }^{ 2 }+{ MR }^{ 2 } } { K }_{ 2 }=\frac { { UMR }^{ 2 } }{ { mr }^{ 2 }+{ MR }^{ 2 } }$
0%
$\displaystyle U=\frac { { \mu }_{ 0 }\pi { I }^{ 2 }{ r }^{ 2 } }{ 4R }$
0%
$\displaystyle { K }_{ 2 }>>{ K }_{ 1 }$

Explanation

	Coil 1	Coil 2
Radius	R	r(<<R)
Current	I	I/2
Mass	M	m
Kinetic energy	$K_1$	$K_2$

$K.E=\dfrac{1}{2}Iw^2$ (1)

$\therefore K-1=\dfrac{1}{2}\times Iw^2\implies I\propto\dfrac{1}{K.E}$

${ K }_{ 2 }=\frac { { MR }^{ 2 } }{ 2 } { w }^{ 2 }$

Now

${ { K }_{ 1 }/K }_{ 2 }=\dfrac { m }{ m } \left( \dfrac { r }{ R } \right) ^{ 2 }$ and

$K_1+K_2=U$ (ii)

$\implies K_2\times\dfrac{m}{M}(r/R)^2+K_2=U$

$\implies K_2=\dfrac{UMR^2}{mr^2+MR^2}$

Similarly

${ K }_{ 1 }=\dfrac { UM{ R }^{ 2 } }{ { mr }^{ 2 }+{ MR }^{ 2 } }$

And using (ii)

$U=U=\dfrac { 1 }{ 2 } \frac { \pi { r }^{ 2 }{ \mu }_{ 2 }I }{ 2R } =\dfrac { { \mu }_{ 2 }{ I }^{ 2 }\pi { r }^{ 2 } }{ 4R } =U$

If you place a compass near a current conducting wire, it will:

0%
Get deflected
0%
Get charged
0%
Get hot
0%
Glow brightly

A particle of charge per unit mass $\alpha$ is released from origin with a velocity $\mathop v\limits^ \to = {v_0}\mathop i\limits^ \wedge$ in a uniform magnetic field $\mathop B\limits^ \to = - {B_0}\mathop k\limits^ \wedge$ . If the particle passes through (0,y,0) then y is
equal to

0%
$- \frac{{2{v_0}}}{{{B_0}\alpha }}\,\;$
0%
$\frac{{{v_0}}}{{{B_0}\alpha }}\;$
0%
$\;\frac{{2{v_0}}}{{{B_0}\alpha }}$
0%
$- \frac{{{v_0}}}{{{B_0}\alpha }}$

Magnetic field at the centre of regular polygon of

$'n'$ sides which is formed by wire, which carries current

$I$ and side of polygon is a:

0%
$\dfrac { n \mu _ { 0 } I } { \pi a } \sin \left( \dfrac { \pi } { n } \right) \cot \left( \dfrac { \pi } { n } \right)$
0%
$\dfrac { n \mu _ { 0 } I } { \pi a } \cos \left( \dfrac { \pi } { n } \right) \tan \left( \dfrac { \pi } { n } \right)$
0%
$\dfrac { n \mu _ { 0 } I } { \pi a } \sin \left( \dfrac { \pi } { n } \right) \tan \left( \dfrac { \pi } { n } \right)$
0%
$\dfrac { n \mu _ { 0 } I } { \pi a }$

A straight magnetised wire of magnetic moment

$10 A {m}^{2}$ is bent into a semi circle. The decrease in the magnetic moment is

0%
$10 \left(\dfrac{\pi - 2}{\pi}\right) A {m}^{2}$
0%
$\dfrac{20}{\pi} A {m}^{2}$
0%
$\dfrac{5}{\pi} A {m}^{2}$
0%
$5 \pi A {m}^{2}$

Explanation

We know that the length of a circular path is

$2\pi r$ where r is its radius.

So the length(L) of a semi circle is

$L=\pi r.$

If the wire is bent into a semi circle, then the straight distance(L') between the poles is the diameter i.e.

$L'=2r.$

Thus

$L'=\cfrac{2L}{\pi}$

The magnetic moment is directly proportional to the straight distance between the poles.

$\therefore\cfrac {M'}M=\cfrac{L'}L=\cfrac2{\pi}$

$\cfrac{M'-M}M=\cfrac{2-\pi}{\pi}$

$\therefore M'-M=M\times{\cfrac{2-\pi}{\pi}}$

But the change in moment

$=M-M'=M\times{\cfrac{\pi-2}{\pi}} =10\times{\cfrac{\pi-2}{\pi}}Am^2$

A coil carrying current

$l$ has radius r and number of units n. It is rewound so that radius of new coil is

$\dfrac{r}{4}$ and it carries current

$l$ . The ratio of magnetic moment of new coil to that of original coil is

0%
$1$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{8}$

Explanation

Radius and number of turns of the coil initially are

$r$ and

$n$ respectively.

Thus initial length of the wire,

$L = n (2\pi r)$

Area of the coil,

$A = \pi r^2$

Initial magnetic moment,

$\mu = n l A = n l \pi r^2$ ..........(1)

Now the same wire is rewound to form the coil of radius

$r' = \dfrac{r}{4}$

Let the number of turns in the new coil be

$n'$

Length of the wire,

$L' = n' (2\pi r') = \dfrac{2\pi r}{4} n'$

But

$L = L'$ , we get,

$2\pi r n = \dfrac{2\pi r}{4}n'$

$\implies n' = 4n$

Area of the new coil,

$A' = \pi (r')^2 = \dfrac{\pi r^2}{16}$

Magnetic moment of the new coil,

$\mu' = n' l A' = 4n l \times \dfrac{\pi r^2}{16} = \dfrac{nl \pi r^2}{4}$

$\therefore$

$\mu' = \dfrac{\mu}{4}$

$\implies \dfrac{\mu'}{\mu} = \dfrac{1}{4}$

The potential energy of a bar magnet of magnetic moment

$M$ placed in a magnetic field of induction

. The position of stable equilibrium of the magnet is at the angular position given by

$\theta$ equal to:

0%
$0^0$
0%
$90^0$
0%
$45^0$
0%
$180^0$

Explanation

For stable equilibrium magnet field

$\vec{B}$ and magnetic moment

$'M'$ should be in parallel then angle

$\theta=0^\circ$ .

Two long straight wires are placed along x-axis and y-axis. They carry current

$I_1$ and

$I_2$ respectively. The equation of the locus of zero magnetic induction in the magnetic field produced by them is?

0%
$y=x$
0%
$y=\left(\displaystyle\frac{I_2}{I_1}\right)x$
0%
$\displaystyle y=\left(\displaystyle\frac{I_1}{I_2}\right)x$
0%
$y=(I_1I_2)x$

Explanation

magnetic field

$B=\dfrac{μ_0I}{2πr}$

the magnetic field at any point in the xy plane

$B=\dfrac{μ_0I_1}{2πy}-\dfrac{μ_0I_2}{2πx}$

( assuming current flowing in x and y-direction respectively.)

So, the equation of the locus of zero magnetic induction,

$\dfrac{I_1}{y}=\dfrac{I_2}{x}$

$y=\dfrac{I_1x}{I_2}$

An electron accelerated through a potential difference

$V$ passes through a uniform transverse magnetic field and experiences a force

$F$ . If the accelerating potential is increased to

$2V$ , the electron in the same magnetic field will experience a force :

0%
$F$
0%
$F/2$
0%
$\sqrt {2} F$
0%
$2F$

Explanation

force is given as:

$\quad F=Bqv\quad$

The kinetic energy of an electron after passing through a potential difference of V volt is given as

But

$\cfrac { 1 }{ 2 } m{ v }^{ 2 }=eV$ or

$V=\sqrt { \cfrac { 2eV }{ m } }$

$F=Bq\sqrt { \cfrac { 2eV }{ m } } \Rightarrow F\propto \sqrt { V } ;F'\propto \sqrt { 2V } ;$

$\cfrac { F' }{ F } =\sqrt { 2 } \quad or\quad F'=\sqrt { 2 } F$

$AB$ is a long wire carrying a current

${I}_{1}$ and

$PQRS$ is a rectangular loop carrying currect

${I}_{2}$ (as shown in the figure)
which among the following statements are correct?
(a) Arm

$PQ$ will get attracted to wire

$AB$ , and the arm

$RS$ will get repelled from wire

$AB$
(b) Arm

$PQ$ will get repelled from wire

$AB$ and arm

$RS$ attracted to wire

$AB$
(c) Forces on the arm

$PQ$ and

$RS$ will be unequal and opposite
(d) Forces on the arm

$PQ$ and

$RS$ will be zero

0%
Only (a)
0%
(b) and (d)
0%
(a) and (c)
0%
(b) and (c)

Explanation

$\because$ current in wire

$AB$ and

$PQ$ are in the same direction. Therefore they will attract

$\because$ current in wire

$AB$ and

$RS$ are in opposite direction. Therefore they will repeal each other
magnetic field around

$PQ$ more than

$RS$ . So force on the arms

$PQ$ and

$RS$ will be unequal and opposite
1144647_721067_ans_12bcc63254744705ab3514c66916734b.png

Two infinitely long straight wires lie in the

$xy$ -plane along the lines

$x = \pm R$ . The wire located at

$x = +R$ carries a constant current

$I_1$ and the wire located at

$x =-R$ carries a constant current

$I_2$ . A circular loop of radius

$R$ is suspended with its centre at

$(0, 0, \sqrt{3}R)$ and in a plane parallel to the

$xy$ -plane. This loop carries a constant current

$I$ in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the

$+\hat{j}$ direction. Which of the following statements regarding the magnetic field

$\overrightarrow{B}$ is (are) true ?

0%
If $I_1 = I_2$ , then $\overrightarrow{B}$ cannot be equal to zero at the origin $(0, 0, 0)$
0%
If $I_1 > 0$ and $I_2 < 0$ , then $\overrightarrow{B}$ can be equal to zero at the origin $(0, 0, 0)$
0%
If $I_1 < 0$ and $I_2 > 0$ , then $\overrightarrow{B}$ can be equal to zero at the origin $(0, 0, 0)$
0%
If $I_1 = I_2$ , then the $z$ -component of the magnetic field at the centre of the loop is $\left(-\dfrac{\mu_0 I}{2R}\right)$

Explanation

(A) At origin,

$\overrightarrow{B} = 0$ due to two wires if

$I_1 = I_2$ (they cancel each other)

hence

$(\overrightarrow{B}_{net})$ at origin is equal to

$\overrightarrow{B}$ due to ring, which is non-zero.
(B) If

$I_1 > 0$ and

$I_2 < 0, \overrightarrow{B}$ at origin due to wires will be along

$+\hat{k}$ direction and

$\overrightarrow{B}$ due to ring is along

$-\hat{k}$ direction and hence

$\overrightarrow{B}$ can be zero at origin.
(C) If

$I_1 < 0$ and

$I_2 > 0, \overrightarrow{B}$ at origin due to wires is along

$-\hat{k}$ and also along

$-\hat{k}$ due to ring, hence

$\overrightarrow{B}$ cannot be zero.
(D) (ref. image 2) At centre of ring,

$\overrightarrow{B}$ due to wires is along

$x$ -axis,
hence

$z$ -component is only because of ring which

$\overrightarrow{B} = \dfrac{\mu_0I}{2R} (-\hat{K})$

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is

$B_1$ . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is

$B_2$ . The ratio

$\displaystyle\frac{B_1}{B_2}$ is?

0%
$\sqrt{3}$
0%
$\displaystyle\frac{1}{\sqrt{2}}$
0%
$2$
0%
$\sqrt{2}$

Explanation

The dipole moment of circular loop is m

$m_1= I.A = I. \pi R^2$ { R = radius of the loop}

$B_1 = \dfrac{\mu_{o}I}{2R}$

moment becomes double, R becomes

$\sqrt{2}R$ (keeping the current constant)

$m_2 = I. \pi(\sqrt{2}R)^2= 2. I \pi R^2= 2 m_1$

$B_2 = \dfrac{\mu_o I}{2 \sqrt{2}R}=\dfrac{B_1}{\sqrt{2}}$

$\dfrac{B_1}{B_1}= \sqrt{2}$

A short bar magnet of magnetic moment m=

$0.32JT^1$ is placed in a uniform magnetic field of

$0.15 T$ . If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

0%
(a) $\overrightarrow {m}$ parallel to $\overrightarrow{B}= -mB = -11.8 \times 10^{-2}J:$ unstable.
(b) $\overrightarrow {m}$ anti-parallel to $\overrightarrow {B}:+mB = +12.8 \times 10^{-2}J$ ; stable.
0%
(a) $\overrightarrow {m}$ parallel to $\overrightarrow{B}= -mB = -4.8 \times 10^{-2}J:$ stable.
(b) $\overrightarrow {m}$ anti-parallel to $\overrightarrow {B}:+mB = +4.8 \times 10^{-2}J$ ; unstable.
0%
(a) $\overrightarrow {m}$ parallel to $\overrightarrow{B}= -mB = -4.8 \times 10^{-2}J:$ unstable.
(b) $\overrightarrow {m}$ anti-parallel to $\overrightarrow {B}:+mB = +4.8 \times 10^{-2}J$ ; stable.
0%
(a) $\overrightarrow {m}$ parallel to $\overrightarrow{B}= -mB = -9.8 \times 10^{-2}J:$ stable.
(b) $\overrightarrow {m}$ anti-parallel to $\overrightarrow {B}:+mB = +10.8 \times 10^{-2}J$ ; unstable.

Two identical conducting wires

$AOB$ and

$COD$ are placed at right angles to each other. The wire

$AOB$ carries an electric current

$I_{1}$ and

$COD$ carries a current

$I_{2}$ . The magnetic field on a point lying at a distance

$'d'$ from

$O$ , in a direction perpendicular to the plane of the wires

$AOB$ and

$COD$ , will be given by

0%
$\dfrac { { \mu }_{ 0 } }{ 2\pi } \left( \dfrac { { I }_{ 1 }+{ I }_{ 2 } }{ d } \right) ^{ 1/2 }$
0%
$\dfrac { { \mu }_{ 0 } }{ 2\pi d } \left( { I }_{ 1 }^{ 2 }+{ I }_{ 2 }^{ 2 } \right) ^{ 1/2 }$
0%
$\dfrac { { \mu }_{ 0 } }{ 2\pi d } \left( { I }_{ 1 }+{ I }_{ 2 } \right)$
0%
$\dfrac { { \mu }_{ 0 } }{ 2\pi d } \left( { I }_{ 1 }^{ 2 }+{ I }_{ 2 }^{ 2 } \right)$

The work done in moving a dipole from its most stable to most unstable position in a

$0.09$

$T$ uniform magnetic field is?(dipole moment of this dipole

$=0.5$

$A$

$m^2$ ).

0%
$0.07 J$
0%
$0.08 J$
0%
$0.09 J$
0%
$0.1 J$

Explanation

Since the most stable position is at

$\theta =0$ and the most unstable position is at

$\theta =180^o$ , then the work done is given by

$W=\displaystyle\int^{\theta =180^o}_{\theta =0^o} \tau{\theta}d\theta=\displaystyle\int^{180^o}_{0^o} MB\sin\theta d\theta =-MB[\cos\theta]^{180^o}_0$

$=-MB[\cos 180^o-\cos 0^o]=-MB[-1-1]$

$=-MB[-2]=2$ MB
Here,

$M=0.5A m^2$ and

$B=0.09T$

$W=2\times 0.50\times 0.09=0.09$ J

The magnetic field due to current flowing in a ling straight conductor is directly proportional to the current and inversely proportional to the distance of the point of observation from the conductor. What is this law known as?

0%
Blonde-Rey law
0%
Biot-Savart's law
0%
Beer-Lambert law
0%
Ampere's law

An electron gun G emits electrons of energy

$2$ keV travelling in the positive x-direction. The electrons are required to hit the spot S where GS

$=0.1$ m, and the line GS makes an angle

$60^o$ with the x-axis, as shown. A uniform magnetic field B parallel to GS exists in the region outside the electron gun. Find the minimum value of B needed to make the electrons hit S.

0%
$1.7\times 10^{-3}$ T
0%
$5.7\times 10^{-3}$ T
0%
$3.7\times 10^{-3}$ T
0%
$4.7\times 10^{-3}$ T

As shown in the figure, four identical loops are placed in a uniform magnetic field B. The loops carry equal current i.

$\hat{n}$ denotes the normal to the plane of each loop. Potential energies in descending order are.

0%
I, II, III, IV
0%
IV, II, III, I
0%
I, III, II, IV
0%
IV, III, II, I

Magnetic field at point

$'P'$ due to given current distribution is:

0%
$\dfrac{{{\mu _0}I}}{{4\pi r}}\left( {1 + \sqrt 2 } \right) \odot$
0%
$\dfrac{{{\mu _0}I}}{{2\pi r}}\left( {1 + \sqrt 2 } \right) \odot$
0%
$\dfrac{{{\mu _0}I}}{{4\pi r}}\left( {1 + \sqrt 2 } \right) \otimes$
0%
$\dfrac{{{\mu _0}I}}{{2\pi r}}\left( {1 + \sqrt 2 } \right) \otimes$

Explanation

$= \dfrac{{{\mu _0}I}}{{4\pi r}}\left( {\sin {{90}^o} + \sin {{45}^o} + \sin {{45}^o}} \right)$

$= \dfrac{{{\mu _0}I}}{{4\pi r}}\left( {1 + \sqrt 2 } \right) \odot$

An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?

0%
The electron will be accelerated along the axis.
0%
The electron path will be circular about the axis.
0%
The electron will experience a force at 45 to the axis and hence execute a helical path.
0%
The electron will continue to move with uniform velocity along the axis of the solenoid.

Explanation

Hint:- If a charge $q$ is moving with velocity $v$ in a Magnetic Field $B$ , such that angle between velocity of charge and Magnetic Field is $\theta$ . Then Force acting on charge is given by,

$F = qvBsin\theta$

Step 1: Note the given values and calculate force on an electron in magnetic field.

$\bullet$ It is given that an electron is projected with uniform velocity along the axis of current carrying long solenoid. Let charge on electron is

$e$ and uniform velocity of electron is

$v$ . Also let at any point on axis of solenoid Magnetic Field is

$B$ .

$\bullet$ We know that Magnetic Field on axis of solenoid is always parallel to the axis. And since electron is moving along axis of solenoid, we have

$\theta = 0°$ or

$\theta = 180°$ . In both the cases value of

$\sin\theta$ is zero.

$\bullet$ Thus net force acting on electron, that is

$F = qvBsin\theta = 0$

Step 2: Explanation for options:-

$\bullet$ Since net force on electron is zero, by Newton's Second Law its acceleration will be zero. Option A is incorrect.

$\bullet$ Also since net force on electron is zero, its direction of motion will not change. So Option B is incorrect.

$\bullet$ As theta is zero not 45 so option C is incorrect.

$\bullet$ As net force is zero so net acceleration is zero. So the electron will continue to move along the axis with same velocity.

Thus, only Option D is correct.

A neutron, a proton, an electron and an

$\alpha$ particle enters a uniform magnetic field with equal velocities. The field is directed along the inward normal to the plane of the paper. Which of these tracks followed are by electron and

$\alpha$ particle ?

0%
A
0%
B
0%
C
0%
D

$\alpha$ - particle crosses a space without any deflection. If electric field E = 8 x

$10^{6}$ V/m and magnetic field is B = 1.6 T, the velocity of particle is

0%
(a) 2.5 x $10^{6}$ m/s
0%
(b) 5 x $10^{6}$ m/s
0%
(c) 8 x $10^{6}$ m/s
0%
(d) 5 x $10^{7}$ m/s

A particle of charge +q and mass m moving under the influence of a uniform electric field

$E\hat{i}$ and a uniform magnetic field

$B\hat{k}$ follows trajectory from P to Q as shown in figure. The velocities at P and Q are

$V\hat{i}$ and

$-2V\hat{j}$ respectively. Which of the following statement(s) is/are correct?

0%
$E= \dfrac{3mv^2}{4 qa}$
0%
Rate of work done by electric field at P is $\dfrac{3mv^3}{4a}$
0%
Rate of work done by electric field at P is zero.
0%
Rate of work done by both the fields at Q is zero.

Explanation

In going from P to Q increase in kinetic energy

$\dfrac{1}{2}m(2V)^2-\dfrac{1}{2}mV^2= \dfrac{1}{2}m(3V^2)=$ Work done by electric field
or,

$\dfrac{3}{2}mv^2= Eq\times 2a \Rightarrow E= \dfrac{3mv^2}{4qa}$
The rate of work done by E at P = force due to E \times velocity.

$= qEa= q\left(\dfrac{3mv^2}{4qa}\right)v = \dfrac{3mv^3}{4a}$
At q,

$\overrightarrow{v}$ is perpendicular to

$\overrightarrow {E}$ and

$\overrightarrow {B}$ , therefore no work done by the either field. Hence (A), (B) & (D) are correct.

A triangular loop of side

$l$ carries a current

$i$ . It is placed in a magnetic field B such that the plane of the loop is in the direction of

$B$ . The torque on the loop is:

0%
$iBl$
0%
${i}^{2}Nl$
0%
$\cfrac { \sqrt { 3 } }{ 4 } Bi{ l }^{ 2 }$
0%
infinity

Explanation

Triangular loop of side having length=L and magnetic field=B

When current(i) is passed through the loop having area (A), then the magnetic dipole moment is given by,

$M=iA$ where, A is the area of the coil

$A=\dfrac{1}{2}\times base\times height=\dfrac{1}{2}\times l\times l\sin60=\dfrac{\sqrt3}{4}l^2$

Therefore magnetic dipole moment is,

$M=iA=\dfrac{\sqrt3}{4}il^2$

The angle between the normal of the plane and the applied magnetic field is 90 degree

Now the torque acting on the loop is,

$\tau=MB\sin\theta=\dfrac{\sqrt3}{4}il^2B\sin90=\dfrac{\sqrt3}{4}iBl^2$

If the radius of the circle is 0.5 m and the magnitude of the magnetic field is

$1.2 W bm^{-2}$ , find the frequency and the kinetic energy of the proton. Charge of the proton =

$1.60\times 10^{-19}C.$ Mass of the proton =

$1.67\times 10^{-27}kg$ .

0%
$1.83\times 10^7Hz, 2.76 \times 10^{-12}J$
0%
$1.83\times 10^7Hz, 3.76 \times 10^{-12}J$
0%
$2.83\times 10^7Hz, 2.76 \times 10^{-12}J$
0%
$2.83\times 10^7Hz, 3.76 \times 10^{-12}J$

A charged particle

$q$ enters a region of uniform magnetic field (out of the page) end is deflected a distance

$d$ after travelling a horizontal distance

$a$ . The magnitude of the momentum of the particle is:

0%
Not possible to be determined as it keeps changing.
0%
$\dfrac{qB}{2}\left[\dfrac{a^{2}}{d}+d\right]$
0%
$\dfrac{qB}{2}$
0%
Zero

Explanation

The vector product between

$v$ and

$B$ points upwards in the figure thus indicating that the charge of the particle is positive.

The Force acting on the moving charge

$F = qvB$ .

As a result of the magnetic force, the charged particle will follow a spherical trajectory.

The centripetal force

${F_c} = \dfrac{{m{v^2}}}{r}$ .

Force acting on the moving charge

$F = F_c$ ,

$\ qvB = \dfrac{{m{v^2}}}{r}$

$r = \dfrac{{mv}}{{qB}} = \dfrac{p}{{qB}}$ ,where

$p$ is the momentum of the particle.

$\begin{array}{l} { r^{ 2 } }={ d^{ 2 } }+{ \left( { r-a } \right) ^{ 2 } } \\ =\dfrac { { { d^{ 2 } }+{ a^{ 2 } } } }{ { 2d } } \\ =\dfrac { 1 }{ 2 } \left( { d+\dfrac { { { a^{ 2 } } } }{ d } } \right) \end{array}$

So,

$\begin{array}{l} p=qBr \\ =\dfrac { { qB } }{ 2 } \left( { d+\dfrac { { { a^{ 2 } } } }{ d } } \right) \end{array}$

A beam of electrons moving along +y direction enters in a region of uniform electric and magnetic fields. If the beam goes undeflected through this region then field (B) and (E) are directed respectively:

0%
-y axis and -z axis
0%
+z axis and +xaxis
0%
+ x axis and -x axis
0%
- x axis and -y axis

Two particle, A and B having equal mass from being accelerated through the same potential difference enter a region of uniform magnetic field and particle describe a circular path of radii

$R_1$ and

$R_2$ respectively. The ratio of the charge on A and B is:

0%
$\lgroup \frac{R_2}{R_1} \rgroup^{\frac{1}{2}}$
0%
$\frac{R_2}{R_1}$
0%
$\lgroup \frac{R_2}{R_1} \rgroup^2$
0%
$\lgroup \frac{R_1}{R_2} \rgroup^2$

The directions of the magnetic field due to the dipole are opposite at:-

0%
$P_1$ and $P_2$
0%
$Q_1$ and $Q_2$
0%
$P_1$ and $Q_1$
0%
$P_2$ and $Q_2$

Two particles of charges +Q and -Q are projected from the same point with a velocity v in a region of uniform magnetic field B such that the velocity vector makes an angle

$\theta$ with the magnetic field.Their masses are M and 2M, respectively.Then,they will meet again for the first time at a point whose distance from the point of projection is

0%
$\dfrac { 2\pi Mvcos\theta }{ QB }$
0%
$\dfrac { 8\pi Mvcos\theta }{ QB }$
0%
$\dfrac { \pi Mvcos\theta }{ QB }$
0%
$\dfrac { 4\pi Mvcos\theta }{ QB }$

A particle of charge

$(5\ C)$ and mass

$10\ g$ moves in a circular orbit of radius (

$2\ cm$ ) with angular speed (

$5\ rad/s$ ). The ratio of the magnitude of its magnetic moment to that of its angular momentum is

0%
$500\ C/kg$
0%
$1000\ C/kg$
0%
$250\ C/kg$
0%
$1\ C/kg$

A disc of radius r and carrying positive charge q is rotating with an angular speed

$\omega$ in a uniform magnetic field B about a fixed axis as shown in the figure, such that angle by the axis of the disc with the magnetic field is

$\theta$ . The torque applied by axis on the disc is

0%
a) $\frac{ q \omega r^2 B \ sin \theta}{2} , clockwise$
0%
(b) $\frac{{q\omega {r^2}Bsin\theta }}{4},anticlockwise$
0%
c) $\frac{ q \omega r^2 B \ sin \theta}{2} , anticlockwise$
0%
d) $\frac{ q \omega r^2 B \ sin \theta}{4} , clockwise$

An electron moves along a straight line inside a charged parallel plate capacitor of uniform surface charge density

$\sigma$ .The space between the plates is filled with constant magnetic field of induction 'B' as shown.Time taken by electron to cross the capacitor is

0%
$\cfrac { e\sigma }{ { \varepsilon }_{ 0 }lB }$
0%
$\cfrac { { \varepsilon }_{ 0 }lB }{ \sigma }$
0%
$\cfrac { e\sigma }{ { \varepsilon }_{ 0 }B }$
0%
$\cfrac { { \varepsilon }_{ 0 }B }{ e\sigma }$

A long parallel straight conductors carry

$i_{1}$ and

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

$20 \mu T$ . If the direction of

$i_{2}$ is reversed, the field becomes

$50 \mu T.$ The ratio of the current

$i_{1}/ i_{2}$ is:

0%
$\dfrac{5}{2}$
0%
$\dfrac{7}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{5}{3}$

The coil of a dynamo is rotating in a magnetic field. The developed induced emf changes and the number of magnetic lines of force also changes. Which of the following conditions is correct?

0%
Lines of force minimum but induced emf is zero
0%
Lines of force maximum but induced emf is zero
0%
Lines of force maximum but induced emf is not zero
0%
Lines of force maximum but induced emf is also maximum

The radiation corresponding to

$3\rightarrow 2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter magnetic field of

$3\times10^{-4}$ . If the radius of the largest circular path followed by these electron is 10.0 mm, the work function of the metal is close to :

0%
0.8 eV
0%
1.6 eV
0%
1.8 eV
0%
1.1 eV

A wire carrying current I has the shape as shown in adjoining figure.
Linear parts of the wire are very long and parallel to X-axis while semicircle portion of radius R is lying Y-Z plane. Magnetic feild at point O is

0%
$\vec B = - \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i + 2\hat k} \right)$
0%
$\vec B = \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i - 2\hat k} \right)$
0%
$\vec B = \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i - 2\hat k} \right)$
0%
$\vec B = - \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i + 2\hat k} \right)$

A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetization of

$5.30\times10^3 A/m$ .
What is its magnetic dipole moment?

0%
$1\times10^{-2 }J/T$
0%
$2.08\times10^{-2} J/T$
0%
$3.08\times10^{-2} J/T$
0%
$1.52\times10^{-2} J/T$

A charge

$q$ is spread uniformly over an insulated loop of radius

$r$ . If it is rotated with an angular velocity

$\omega$ with respect to normal axis then magnetic moment of the loop is

0%
$\dfrac {3}{2} q\omega r^{2}$
0%
$\dfrac {1}{2} q\omega r^{2}$
0%
$q\omega r^{2}$
0%
$\dfrac {4}{3}q\omega r^{2}$

A charged particle of charge q and mass, gets deflected through an angle

$\theta$ upon passing through a square region of side 'a' which contains a uniform magnetic field B normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?

0%
$\dfrac { qB }{ m } a cot \theta$
0%
$\dfrac { qB }{ m } a tan \theta$
0%
$\dfrac { qB }{ m } a cot^2 \theta$
0%
$\dfrac { qB }{ m } a tan^2 \theta$

A current of i ampere is flowing in an equilateral triangle of side a. The magnetic induction at the centroid will be?

0%
$\dfrac{\mu_i}{3\sqrt{3}\pi a}$
0%
$\dfrac{3\mu_i}{2\pi a}$
0%
$\dfrac{5\sqrt{2}\mu_i}{3\pi a}$
0%
$\dfrac{9\mu_i}{2\pi a}$

Explanation

$\textbf {Hint :}$ Use magnetic field due to finite straight current carrying wire,

$B=\dfrac { μ }{ 4π } \times \dfrac { i }{ r } \times (Sinθ_1+Sinθ_2)$

$\textbf{Step 1: Calculating perpendicular distance}$

Length of side

$=a$

distance

$= GD = \dfrac{1}{3} \times AC \sin 60$

$r= \dfrac{a}{2\sqrt{3}}$

$\textbf{Step 2: Magnetic field due to wire BC}$

Now field at centroid G, due to BC

$B_1 = \dfrac{\mu_0i}{4\pi \dfrac{a}{2\sqrt{3}}} (\sin 60^o + \sin 60^o)$

$B_1 = \dfrac{3\mu _0 i}{2\pi a}$

$\textbf{Step 3: Calculating net magnetic field}$

Similer for all those branches

$B_{net} = 3B_1$

$B_{net} = \dfrac{9\mu_0i}{2\pi a}$

${\textbf{Option D is correct}}$

A electron moves in a circular orbit at a distance from a proton with kinetic energy E. to escape to infinity, the energy which must be supplied to the electron is:

0%
E
0%
$2E$
0%
$0.5E$
0%
$\sqrt2E$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 13 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 13 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.