MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 14

Two long straight wires are arranged in

$x-y$ plane along

$x=\pm d$ . The wires carry current along

$+y$ axis. Another wire of mass per unit length

$\lambda$ carrying current in

$-y$ axis is placed along line

$x=0$ . Currents in all three wires are

$I$ . If the middle wire is displaced slightly , find time period of small oscillations

0%
$2\pi \sqrt { \dfrac { { \lambda d }^{ 2 } }{ { \mu _{ 0 }I }^{ 2 } } }$
0%
$2\sqrt { \dfrac { { \lambda d }^{ 2 } }{ { \pi \mu _{ 0 }I }^{ 2 } } }$
0%
${ 2\pi }^{ 2 }\sqrt { \dfrac { { \lambda d }^{ 2 } }{ { \pi \mu _{ 0 }I }^{ 2 } } }$
0%
${ 2\pi }^{ 2 }\sqrt { \dfrac { { \lambda d }^{ 2 } }{ { \mu _{ 0 }I }^{ 2 } } }$

$\bar B_1,\bar B_2 and \bar B_3$ are the magnetic field due to

$i_1, I_2,and I_3$ then in Ampere's circuital law

$\oint \bar B-d\bar I= \mu_0I, \bar B$ is

0%
$\bar B=\bar B_1-\bar B_2$
0%
$\bar B=\bar B1+\bar B_2+\bar B_3$
0%
$\bar b=\bar B_1-\bar B_2+\bar B_3$
0%
$\bar B=\bar B_3$

Axis of a solid cylinder of infinite length and radius

$R$ lies along y-axis it carries a uniformly distributed current '

$i$ ' along +y direction. Magnetic field at a point

$\left( \dfrac { R } { 2 } , y , \dfrac { R } { 2 } \right)$ is

0%
$\dfrac { \mu _ { 0 } \mathbf { i } } { 4 \pi R } ( \hat { i } - \hat { j } )$
0%
$\dfrac { \mu _ { 0 } i } { 2 \pi R } ( \hat { j } - \hat { k } )$
0%
$\dfrac { \mu _ { 0 } \mathrm { i } } { 4 \pi \mathrm { R } } \hat { \mathrm { j } }$
0%
$\dfrac { \mu _ { 0 } \mathbf { i } } { 4 \pi R } ( \hat { i } + \hat { k } )$

An electron at point A in figure has a speed

$v_0$ of magnitude 1.41 x

$10^6$ m/s. It enters into a uniform magnitude field and follows a semicircular path in it as shown.
The time required for the electron to move from A to B is

0%
1.11 x $10^{-7}$ s
0%
1.11 x $10^6$ s
0%
1.11 x $10^5$ s
0%
1.11 s

Rank the value of

$\oint{\overset{-}{B}}.dl$ for the paths shown in figure from the smallest to largest:

0%
a, b, c, d
0%
a, c, d, b
0%
a, d, c, b
0%
a, c, b, d

An arc of a circle of radius

$R$ subtends an angle

$\dfrac{\pi}{2}$ at the centre. It carries a current

$i$ . The magnetic field at the centre will :

0%
$\dfrac{\mu_{0}i}{2R}$
0%
$\dfrac{\mu_{0}i}{8R}$
0%
$\dfrac{\mu_{0}i}{4R}$
0%
$\dfrac{\mu_{0}i}{5R}$

Explanation

Magnetic field due to current carring arc

$dB=\dfrac{\mu_o}{4\pi}\dfrac{idl\sin 90}{R^2}$

$=\dfrac{\mu_o}{4\pi}\cdot \dfrac{idl}{R^2}$

$N_{Net}=\displaystyle\int dB=\displaystyle\int \dfrac{\mu_o}{4\pi}\dfrac{idl}{R^2}$

$\left\{\theta =\dfrac{l}{R}\right\}$

$B_{Net}=\dfrac{\mu_o}{4\pi}\dfrac{i}{R^2}\displaystyle\int dl$

$=\dfrac{\mu_o}{4\pi}\dfrac{i}{R^2}(R\theta)$

$B_{net}=\dfrac{\mu_o}{4\pi}\cdot \dfrac{i\theta}{R}$

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

$B_{Net}=\dfrac{\mu_o}{4\pi}\cdot \dfrac{i\times \pi/2}{R}$

$=\dfrac{\mu_oi}{8R}$ .

1445366_1210924_ans_ed86b89a1c0f44978b5dc42e850bc0dc.png

A charged particle of charge 5 mc and mass 5 gm is moving with a constant speed 5 m/s in a uniform magnetic field on a curve

$x^2 + y^2 = 25$ . Where x and y are in meter. The value of magnetic field required will be

0%
1 Tesla
0%
1 T along z-axis
0%
5 KT along the x-axis
0%
1 KT along any line in the x-y plane

$25cm$ long solenoid has radius

$2cm$ and

$500$ total number of turns. It carries a current of

$15A$ . If it is equivalent to a magnet of the same size and magnetization

$\overrightarrow { M }$ (magnetic moment/volume), then

$\left| \overrightarrow { M } \right|$ is:

0%
$3000\pi A{m}^{-1}$
0%
$3\pi A$ ${m}^{-1}$
0%
$30000 A$ ${m}^{-1}$
0%
$300 A$ ${m}^{-1}$

A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of

${75}^{o}$ . One of the fields has a magnitude of

$15mT$ . The dipole attains stable equilibrium at an angle of

${30}^{o}$ with this field. The magnitude of the other field (in

$mT$ ) is close to

0%
$36$
0%
$1060$
0%
$11$
0%
$1$

The resultant magnetic moment for the following arrangement is (the vectors are non coplanar)

0%
M
0%
2 M
0%
3 M
0%
4 M

The magnetic moment of a current carrying loop is

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

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

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

A conducting wire of length

$'\iota '$ is placed on a rough horizontal suriace, wncec a uniform horizontal magnetic field B perpendicular to the length of the wire exists. Least values of the horizontal forces required to move the rod when a current 'I' is established in the rod are observed to be

${ F }_{ 1 }\& { F }_{ 2 }(<{ F }_{ 1 })$ for the two possible directions of the current through the rod respectively. What is the coefficient of friction between the rod and the surface?

0%
$\mu =\dfrac { { F }_{ 1 }-{ F }_{ 2 } }{ BIl }$
0%
$\mu =\dfrac { { F }_{ 1 }+{ F }_{ 2 } }{ BIl }$
0%
$\mu =\dfrac { { F }_{ 1 }+{ F }_{ 2 } }{ 2BIl }$
0%
$\mu =\dfrac { { F }_{ 1 }-{ F }_{ 2 } }{ 2BIl }$

In the figure, a charged sphere of mass

$m$ and charge

$q$ starts sliding from rest on a vertical fixed circular track of radius

$R$ from the position shown. There exists a uniform and constant horizontal magnetic field induction

$B$ . The maximum force exerted by the track on the sphere is :

0%
$mg+qB \sqrt{2gR}$
0%
$3mg-qB \sqrt{2gR}$
0%
$3mg+qB\sqrt2gR$
0%
$mg+qB\sqrt2gR$

Explanation

${F_m} = qvB,$ and directed radially outward

$.$

$\because N - mg\sin \theta + qvB = \dfrac{{m{v^2}}}{R}$

$\Rightarrow N = \dfrac{{m{v^2}}}{R} + mg\sin \theta - qvB$

Hence at

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

$\Rightarrow {N_{\max }} = \dfrac{{2mgR}}{R} + mg - qB\sqrt {2gR}$

$= 3mg - qB\sqrt {2gR}$

Hence,

option

$(B)$ is correct answer.

The number of turns in two concentric coils are ${n_1}\;{\text{and}}\;{n_2}$ and the ratio of their radii is 2:When equal currents is passed through them in opposite direction, the magnetic field produced at the center becomes zero. The ratio ${n_1}:{n_2}$ will be:

0%
1:1
0%
1:2
0%
2:1
0%
None of these

A conducting loop of radius

$a$ and resistance

$R$ is placed in uniform magnetic field perpendicular to its plane. If the bop is rotated by

$180 ^ { \circ }$ about its diameter then the amount of charge flown through the bop will be.

0%
$\frac { \mathrm { \pi a } ^ { 2 } \mathrm { B } } { \mathrm { R } }$
0%
$\frac { 2 \mathrm { \pi a } ^ { 2 } \mathrm { B } } { \mathrm { R } }$
0%
$\frac { \pi a ^ { 2 } B } { 2 R }$
0%
zero

Explanation

Given: radius of loop $=a$

resistance of loop $=R$

loop rotated by $180^{\circ}.$ $.....(1)$

To find: amount of charge flow, $q=?$

Solution: Let time period of loop is

$T.$

On using

$eq^n(1),$ we conclude that time taken to rotate by

$180^{\circ}$ is

$t=T/2.$

$..(1)$

Now, magnetic flux,

$\phi=BAcos(\omega t)$

here

$A$ is the area of loop and

$B$ is magnetic field.

$\phi=B\pi a^2cos(\omega t)$

now e.m.f induced,

$\xi=-\frac{\mathrm{d}\phi}{\mathrm{d}t}$

$==>\xi=-\frac{\mathrm{d}}{\mathrm{d}t} B\pi a^2cos(\omega t)$

$\xi=B\pi a^2\omega sin(\omega t)$

also,

$\xi=IR=R\frac{\mathrm{d}q}{\mathrm{d}t}$

we get,

$R\frac{\mathrm{d}q}{\mathrm{d}t}=B\pi a^2\omega sin(\omega t)$

integrating on both sides w.r.t.

$t$ ,we get

$==>Rq=-B\pi a^2cos(\omega t)$

now

$use$

$eq^n(1)$

$==>Rq=-B\pi a^2 cos(\dfrac{2\pi T}{T*2})$

$==>Rq=-B\pi a^2 cos\pi$

$==>Rq=B\pi a^2$

$==>q=\dfrac{B\pi a^2}{R}$

hence,

The correct opt: A

A small piece of metal

$\left( { \mu }_{ r }=20 \right)$ of volume

$10 c{ m }^{ 2 }$ has a uniform magnetic field 4 T inside it. The magnetic energy stored in the metal is:

0%
3.18 J
0%
5.64 J
0%
6.36 J
0%
1.59 J

Suppose an isolated north pole is kept at the centre of a circular loop carrying a electric current i. The magnetic field due to the north pole at a point on the periphery of the wire is B. The radius of the loop is a. The force on the wire is :-

0%
Nearly $2\pi aiB$ perpendicular to the plane of the wire
0%
$2\pi aiB$ into the plane of the wire
0%
$\pi aiB$ along the axis of the wire
0%
zero

At what distance on the axis, form the center of a circular current carrying coil of radius

$r$ , the magnetic field becomes

$1/8th$ of the magnetic field at center?

0%
$\sqrt {2r}$
0%
$2^ {3/2}r$
0%
$\sqrt {3}r$
0%
$3\sqrt {2r}$

The magnetic field at centre '

$O$ ' of the arc in figure is

0%
$\frac{{{\mu _o}I}}{{8 \times r}}$
0%
$\frac{{{\mu _o}I}}{{2\pi \times r}}\left[ {\frac{\pi }{4} + \left( {\sqrt 2 - 1} \right)} \right]$
0%
$\frac{{{\mu _o}I}}{{4\pi \times r}}\left[ {\sqrt 2 - \pi } \right]$
0%
$\frac{{{\mu _o}I}}{{2 \times r}}$

A magnetic field

$4\times { 10 }^{ -3 }\left( \hat { k } \right)$ T exerts a force

$\left( 4\hat { i } +3\hat { j } \right) \times { 10 }^{ -10 }$ N on a particle having a charge

${ 10 }^{ -9 }C$ and going in the X-Y plane. The velocity of the particle (in m/s) is

0%
$75\vec { i } +100\vec { j }$
0%
$-75\vec { i } +100\vec { j }$
0%
$75\vec { i } -100\vec { j }$
0%
$-75\vec { i } -100\vec { j }$

The magnetic induction field at the centre

$C$ of the wire which is in the shape as shown in figure carrying current '

$i$ ' is

0%
$\frac{{{\mu _o}i}}{{4\pi r}}\left( {1 + \pi } \right)$
0%
$\frac{{{\mu _o}i}}{{2\pi r}}\left( {1 + \pi } \right)$
0%
$\frac{{{\mu _o}i}}{{\pi r}}\left( {1 + \pi } \right)$
0%
$\frac{{{\mu _o}i}}{{ r}}\left( {1 + \pi } \right)$

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it move in a straight path, then the mass of the particle is ___?
(Given charge of electron =

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

0%
$2.0 \times 10^{-24} kg$
0%
$1.6 \times 10^{-19} kg$
0%
$1.6 \times 10^{-27} kg$
0%
$9.1 \times 10^{-31} kg$

Explanation

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

$mv = q BR$
Path is straight line
it qE = qvB
E = vB
From equation (i) & (ii)

$m = \dfrac{qB^2R}{E}$

$m = 2.0 \times 10^{-24} kg$

A bar magnet of magnetic moment

$m$ and moment of inertia

$I$ (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let

$T$ be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field

$B$ .What would be the similar period

$T^ { \prime }$ for each piece?

0%
$T ^ { \prime } = \frac { 1 } { 2 } T$
0%
$T ^ { \prime } = \frac { 1 } { 4 } T$
0%
$T ^ { \prime } = 2 T$
0%
$T ^ { \prime } = T$

Explanation

$Given:$ Magnetic moment=

$m$ ; moment of inertia=

$I$

$Solution:$ Time period of this type of S.H.M is

$T =$

$2 \pi \sqrt{\frac{ I }{ M B }}$

where, I is moment of inertia

$M$ is mass of magnet.

$B$ is magnetic field

According to this problem, a magnet is oscillating in a uniform magnetic field

$so , T =2 \pi \sqrt{\frac{ I }{ M B }}$

Here,

$I =\frac{ m l ^{2}}{ 1 2 }$

when magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to the length passing through its center is,

$I ^{\prime}=\frac{ m ( l / 2 )^{2}}{ 2 \times 4 }=\frac{ I }{ 8 }$

Magnetic dipole moment,

$M ^{\prime}=\frac{ M }{ 2 }$

So, Time period of the oscillation is,

$T ^{\prime}=$

$2 \pi \sqrt{\frac{ I ^{\prime}}{ M ^{\prime} B }}=2 \pi \sqrt{\frac{ I / 8 }{( M / 2 ) B }}=\frac{2 \pi}{ 2 } \sqrt{\frac{ I }{ M B }}$

$\Longrightarrow T ^{\prime}=\frac{ T }{ 2 }$

$So,the$

$correct$

$option:A$

A steel wire of length l has a magnetic moment M. it is then bent into a semicircular arc. the new magnetic moment is

0%
M
0%
$2M \pi^-1$
0%
M/l
0%
$M \times I$

Explanation

$\textbf{Given}$ : Length l and magnetic moment M of steel wire

$\textbf{To find}$ : New magnetic moment

$\textbf{Solution}$ :

We know, pole strength is given by,

m= $\dfrac{M}{l}$

A steel wire is bent into semi-circular arc, so if r is radius of semicircle, then

πr=l,

Therefore, r= $\dfrac{1}{\pi }$

Now new magnetic moment M′= $m\times 2r$

= $\dfrac{M}{l}\times \dfrac{2l}{\pi }$

= $\dfrac{2M}{\pi }$ =2M $\pi ^{-1}$

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

The unit vectors

$\widehat i,$

$\widehat j$ and

$\widehat k$ are as shown below

$.$ What will be the magnetic field at

$O$ in the following figure

0%
$\frac{{{\mu _0}}}{{4\pi }}\frac{i}{a}\left( {2 - \frac{\pi }{2}} \right)\widehat j$
0%
$\frac{{{\mu _0}}}{{4\pi }}\frac{i}{a}\left( {2 + \frac{\pi }{2}} \right)\widehat j$
0%
$\frac{{{\mu _0}}}{{4\pi }}\frac{i}{a}\left( {2 + \frac{\pi }{2}} \right)\widehat i$
0%
$\frac{{{\mu _0}}}{{4\pi }}\frac{i}{a}\left( {2 + \frac{\pi }{2}} \right)\widehat k$

Which of following cannot be deflected by magnetic field

0%
$\alpha -$ rays
0%
$\beta$ rays
0%
$\gamma -$ rays
0%
cosmic rays

Explanation

$\gamma- rays$ will not be deflected by the magnetic field because

$\gamma- rays$ don't have any charge.

Inside a long solenoid wounded with

$300$ turns/ metre, an iron rod is placed. An iron rod is

$0.2\ m$ long,

$10\ mm$ in diameter and of permeability

$10^3$ . The magnetic moment of the rod, if

$0.5\ amp$ of current is passed through the rod, is :

0%
$2.356\ SI\ unit$
0%
$1.335\ SI\ unit$
0%
$3.664\ SI\ unit$
0%
$1.664\ SI\ unit$

A moving coil galvanometer, having a resistance

$G$ , produces full scale deflection when a current

$I_g$ flows through it. This galvanometer can be converted into (i) an ammeter of range

$0$ to

$I_0 (I_0 > I_g)$ by connecting a shunt resistance

$R_A$ to it and (ii) into a voltmeter of range

$0$ to

$V(V = GI_0)$ by connecting a series resistance

$R_V$ to it. Then

0%
$R_A R_V = G^2 \left(\dfrac{I_g}{I_0 - I_g}\right)$ and $\dfrac{R_A}{R_V} = \left(\dfrac{I_0 - I_g}{I_g}\right)$
0%
$R_AR_V = G^2$ and $\dfrac{R_A}{R_V} = \left(\dfrac{I_g}{I_0 - I_g}\right)^2$
0%
$R_AR_V = G^2$ and $\dfrac{R_A}{R_V} = \dfrac{I_g}{I_0 - I_g}$
0%
$R_A R_V = G^2 \left(\dfrac{I_0 - I_g}{I_g}\right)$ and $\dfrac{R_A}{R_V} = \left(\dfrac{I_g}{I_0 - I_g}\right)^2$

Explanation

When galvanometer is used as an ammeter shunt is used in parallel with galvanometer.

$\therefore I_g G = (I_0 - I_g)R_A$

$\therefore R_A = \left(\dfrac{I_g}{I_0 - I_g}\right) G$

When galvanometer is used as a voltmeter, resistance is used in series with galvanometer.

$I_g(G + R_v) = V = GI_0$ (given

$V = GI_0$ )

$\therefore R_v = \dfrac{(I_0 - I_g)G}{I_g}$

$\therefore R_AR_V = G^2 \& \dfrac{R_A}{R_V} = \left(\dfrac{I_g}{I_0 - I_g}\right)^2$
1248173_1614915_ans_79706047d7944b2db7b3d5cb043a6c6c.png

A bar magnet when suspended freely has a time-period of

$4$ second.It is cut into two equal parts along the length. One part is placed over the other such that their centres coincide and their dipole moments make an angle of

600600 with each other.if this assembly is freely suspended, then its time period is.

0%
$2.2 s$
0%
$3.0 s$
0%
$4.3 s$
0%
$6.28 s$

$P, Q$ and

$R$ long parallel straight wires in air carrying currents as shown. The direction of resultant force on

$R$ is

0%
Towards left
0%
Towards right
0%
The same at that current in $Q$
0%
Perpendicular to plane of paper

Explanation

Correct Answer: A

Hint: Calculate net magnetic field direction at R and then use $\vec{F} = i(\vec{dl} \times \vec{B})$

Step-1: Force on R due to the current in P & Q

$\therefore {\text{ Force per unit length on the wire R due to Q wire is }}$

${\text{ }}{{\text{F}}_1} = \dfrac{{{\mu _{o{I_3}{I_2}}}}}{{2\pi r}} = \dfrac{{24{\mu _o}}}{{2\pi r}}{\text{ direction towards left according to Fleming's left hand rule}}$

${\text{ Similarly, force per unit length on the wire R due to P wire is}}$

${\text{ }}{{\text{F}}_2}{\text{ = }}\dfrac{{{\mu _o}{I_1}{I_3}}}{{2\pi \times 2r}}{\text{ = }}\dfrac{{12{\mu _o}}}{{4\pi r}}{\text{ direction towards left according to Fleming's left hand rule}}$

Step-2: Resultant force

$\therefore {\text{ Resultant force }}F{\text{ per unit length }} = {\text{ }}{F_1} + {F_2}$

${\text{ }} = \dfrac{{24{\mu _o}}}{{2\pi r}} + \dfrac{{12{\mu _o}}}{{4\pi r}}$

${\text{ }} = \dfrac{{60{\mu _o}}}{{4\pi r}} = \dfrac{{15{\mu _o}}}{{\pi r}}$

$\because {\text{ }}{{\text{F}}_1}{\text{ and }}{{\text{F}}_2}{\text{ both acting towards left therefore resultant force also acting }}$ ${\text{towards left on the wire R}}$

In an uniform field the magnetic needle completes 10 oscillation in 92seconds. When a small magnet is placed in the magnetic meridian 10 cm due north of needle with north pole towards south completes 15 oscillation in 69 seconds. The magnetic moment of magnet (

$B_H = 0.3 G$ ) is

0%
$4.5 Am^2$
0%
$0.45 Am^2$
0%
$0.75 Am^2$
0%
$0.225 Am^2$

Explanation

Given;

$f_1=\dfrac{10}{92}=0.10869565217 (sec)^{-1}$

$f_2=\dfrac{15}{69}=0.21739130435 (sec)^{-1}$

$B_H=0.3*10^{-4}T$

To find: magnetic moment of magnet,

$M=?$

Solution: As we know that,

$\dfrac{B_2}{B_1}=\dfrac{f_1^2}{f_2^2}$

here,

$B_2$ is field of magnet and

$B_1=B_H$

$==>B_2={B_H}\dfrac{(0.10869565217)^2}{(0.21739130435)^2}$

$...(1)$

also,

$B_2=\dfrac{\mu_0}{4\pi}\dfrac{M}{r^2}$

$...(2)$

equating

$(1),(2)$ we get

$\dfrac{\mu_0}{4\pi}\dfrac{M}{r^2}=0.3*10^{-4}*0.24999999998$

$==>10^{-7}\dfrac{M}{(0.1)^2}=0.07499999999*10^{-4}$

$M=0.75Am^2$

hence,

The correct opt : C

A negative charge is given to a non-conducting loop and the loop is rotated in the plane of paper about its centre as shown in figure. The magnetic field produced by the ring affects a small magnet placed above the ring in the same plane :-

0%
The magnet does not rotate
0%
The magnet rotates clockwise as seen from below
0%
The magnet rotates anticlockwise as seen from below
0%
No effect on magnet is there

The magnetic induction at the point

$O$ . If the wire carries a current

$i$ , is

0%
$\dfrac{\mu_0 i}{2R}$
0%
$\dfrac{\mu_0 i}{2\pi R}$
0%
$\dfrac{\mu_0 i(\pi^2 + 4)^{{1}/{2}}}{4\pi R}$
0%
$\dfrac{\mu_0 i(\pi^2 + 4)}{4\pi R}$

Explanation

Hint: The resultant magnetic field is the vector sum of ${B_x}$ and ${B_y}$ .

Correct Option: C

Explanation for Correct Option:

Step 1: Find the magnetic field due to the half ring wire,

${B_1} = \dfrac{{{\mu _o}i}}{{4R}}$ ,

where $i$ is the current through the wire and $R$ is the radius of the wire.

Step 2: Magnetic Field induction at point O due to the two straight wires,

Magnetic field at O, ${B_2} = 2[\dfrac{{{\mu _o}i}}{{4\pi R}}(\cos 0 - \cos 90)]$

Step 3: Find the resultant magnetic field at point $O$ .

Resultant magnetic field,

$B = \sqrt {{B_1}^2 + {B_2}^2}$

$B = \dfrac{{{\mu _o}i}}{{4\pi R}}{({\pi ^2} + {2^2})^{\dfrac{1}{2}}}$

So, the magnetic induction at point $P$ is $B = \dfrac{{{\mu _o}i}}{{4\pi R}}{({\pi ^2} + {2^2})^{\dfrac{1}{2}}}$

2148340_1595886_ans_269d9eb057044d7dad1e677af55c7dfd.PNG

Two circular concentric loops of radii $r_1=20cm$ and $r_2 =30 cm$ are placed in the X-Y plane as shown in the figure. A current I=7 A is flowing through them. The magnetic moment of this loop system is

0%
$+ 1.1 \hat {k} (Am^2)$
0%
$- 1.5 \hat {k} (Am^2)$
0%
$6$
0%
none of these

Explanation

$\begin{array}{l} Here,\, \, magnetic\, \, moment\, \, due\, \, to\, \, loop\, \, 1 \\ { M_{ 1 } }=i{ A_{ 1 } } \\ =i\pi { r_{ 1 } }^{ 2 } \\ =7\times \pi { \left( { 0.20 } \right) ^{ 2 } }=0.28\pi \\ similarly\, \, magnetic\, \, moment\, \, due\, \, to\, \, loop\, \, 2 \\ { M_{ 2 } }=i{ A_{ 2 } }=i\pi { r_{ 2 } }^{ 2 } \\ =7\times \pi { \left( { 0.30 } \right) ^{ 2 } }=0.63\pi \\ Net\, \, magnetic\, \, moment \\ ={ M_{ 1 } }-{ M_{ 2 } } \\ =0.63\pi -0.28\pi =61.1\, A{ m^{ 2 } }=+1.1\widehat { k } \left( { A{ m^{ 2 } } } \right) \end{array}$

Hence,

option

$(A)$ is correct answer.

An electron gun is placed inside a long solenoid of radius

$R$ on its axis. The solenoid has

$n$ turns / length and carries a current

$I$ . The electron gun shoots an electron along the radius of the solenoid with speed

$\upsilon$ . If the electron does not hit the surface of the solenoid, maximum possible value of

$\upsilon$ is (all symbols have their standard meaning):

0%
$\dfrac {e \mu_{0} nIR}{4m}$
0%
$\dfrac {e \mu_{0} nIR}{m}$
0%
$\dfrac {2e \mu_{0} nIR}{m}$
0%
$\dfrac {e \mu_{0} nIR}{2m}$

Explanation

Magnetic field inside a solenoid,

$B = (\mu_0 nI)$

Since, velocity V is given to the electron.

$q = e =$ charge on an electron.

$m =$ mass of an electron.

Magnetic force,

$F = q \vee B$ Ref. image I

$F = (eV \mu_0 nI)$

In a magnetic field, charge will move on a circular path. Here, If radius is just less than or equal to radius of solenoid then the electron does not hit the surface.

$\therefore F = q B \vee = \dfrac{mV^2}{R}$ Ref. image II

$\Rightarrow mv = q BR$

$\dfrac{mV^2}{R}$ is centripetal force

$\Rightarrow v = \dfrac{(qR)B}{m}$

$q = e$

$\Rightarrow v = \dfrac{qR \times \mu_0 nI}{m} = \dfrac{e \mu_0 nIR}{m}$

$\therefore V = \dfrac{e \mu_0 n IR}{m}$ maximum possible value of V.

Option (2) is correct.

1482527_1697519_ans_42a3a9b0ed2446abbc1fcb7928b0dbed.png

Graph the magnitude of

$\overrightarrow {B}$ at points on the x-axis

Explanation

At position on the x-axis :

$B_{net} = 2 \frac { \mu_0 I}{ 2 \pi r} sin \theta = \frac { \mu_0 I}{ \pi \sqrt { x^2 + a^2 } } \frac {a}{ \sqrt { x^2 +a^2 } }$

$B_{net} = \frac { \mu_0 I a}{ \pi ( x^2 + a^2 ) }$
Its graph given below.
The magnetic field is maximum at origin , x= 0
when x >>a ,

$B \approx \frac { \mu_0 Ia}{ \pi x^2 }$
1620637_1747566_ans_b8899b128ab44688838b61f96dba8800.png

At what value of x is the magnitude of

$\overrightarrow {B}$ is maximum ?

0%
$x = 0$
0%
$x = \sqrt {2} a$
0%
$x = \pm a / 2$
0%
$x = \pm a$

Explanation

At a position on the x-axis :

$B_{net} = 2 \frac { \mu_0 I}{ 2 \pi r } cos \theta = \frac { \mu_0 I}{ \pi \sqrt { x^2 + a^2 } } \frac {x}{ \sqrt { x^2 +a^2 } }$

$B_{net} = \frac { \mu_0 Ix }{ \pi ( x^2 +a^2 ) }$
its graph is shown below.
The magnetic field is maximum when :

$\frac {dB}{dx} = 0 = \frac { C}{ x^2 +a^2 } - \frac { 2Cx^2 }{ (x^2 +a^2 )^2}$

$( x^2 +a^2 ) - 2x^2 \Rightarrow x = \pm a$ .

1620682_1747574_ans_3b11a6eaa9264b078712d7656de66218.png

What is the magnitude of the net field at S ?

0%
$4.1 \times 10^{-6} T$
0%
$1.6 \times 10^{-6} T$
0%
$3.2 \times 10^{-6} T$
0%
$2.1 \times 10^{-6} T$

Explanation

The magnitude of the field at S is given by sum of the square of the two fields because they are right angles So,

$B_S = \sqrt { B^2_1 + B^2_2 } = \frac { \mu_0}{2\pi} \sqrt {(\frac {I_1}{r_1})^2 +(\frac {I_2}{r_2})^2 }$

$\frac { \mu_0}{ 2 \pi } \sqrt { ( \frac {6.00 A}{0.60 m})^2 + ( \frac {2.00 A }{ 0.80 m} ) } = 2.1 \times 10^{-6} T.$

What is the magnitude of the net field at Q ?

0%
$2.13 \times 10^{-6} T$ `
0%
$4.26 \times 10^{-6} T$
0%
$1.21 \times 10^{-6} T$
0%
$5.30 \times 10^{-6} T$

Explanation

The field at Q point to the right and has magnitude

$B_Q = \dfrac { \mu_0}{ 2 \pi} ( \dfrac { I_1}{r_1} - \dfrac {I_2}{r_2 } )$

$= \dfrac { \mu_0}{2 \pi } ( \dfrac { 6.00 A}{0.500 m} - \dfrac { 2.00 A}{1.50 m} )$

$= 2.13 \times 10^{-6} T$

The value of

$\dfrac{\delta B_x}{\delta x}+\dfrac{\delta B_y}{\delta y}+\dfrac{\delta B_z}{\delta z}$ is :

0%
$>0$
0%
$<0$
0%
$>0$
0%
$=0$

Derive the expression for the magnitude of

$\overrightarrow {B}$ at any point on the x-axis in terms of the x-ocoordinate of the point. what is the direction of

$\overrightarrow {B}$ ?

0%
$\frac { \sqrt {2} \mu_0 Ia }{ \pi ( x^2 +a^2 ) }$
0%
$\frac { 3 \mu_0 I a}{ \pi ( x^2 + a^2 ) }$
0%
$\frac { \mu_0 Ia}{ 2\pi ( x^2 + a^2 )}$
0%
$\frac { \mu_ 0 Ix}{ \pi ( x^2 + a^2) }$

Explanation

At a position on the x-axis :

$B_{net} = 2 \frac { \mu_0 I}{ 2 \pi r } cos \theta = \frac { \mu_0 I}{ \pi \sqrt { x^2 + a^2 } } \frac {x}{ \sqrt { x^2 +a^2 } }$

$B_{net} = \frac { \mu_0 Ix }{ \pi ( x^2 +a^2 ) }$
its graph is shown below.
The magnetic field is maximum when :

$\frac {dB}{dx} = 0 = \frac { C}{ x^2 +a^2 } - \frac { 2Cx^2 }{ (x^2 +a^2 )^2}$

$( x^2 +a^2 ) - 2x^2 \Rightarrow x = \pm a$
when x>>a,

$B \approx \frac { \mu_0 I}{ \pi x}$ ,
which is just like wire carrying current 2 I.
1620672_1747572_ans_49b7a045d1284ea88360cb4062c07b83.png

At what value of x , the magnitude of

$\overrightarrow {B}$ is maximum ?

0%
$x = 0$
0%
$x =\sqrt {2} a$
0%
$x = 1 / \sqrt {2} a$
0%
$x = a / 2$

Explanation

At position on the x-axis :

$B_{net} = 2 \frac { \mu_0 I}{ 2 \pi r} sin \theta = \frac { \mu_0 I}{ \pi \sqrt { x^2 + a^2 } } \frac {a}{ \sqrt { x^2 +a^2 } }$

$B_{net} = \frac { \mu_0 I a}{ \pi ( x^2 + a^2 ) }$
Its graph given below.
The magnetic field is maximum at origin , x= 0
when x >>a ,

$B \approx \frac { \mu_0 Ia}{ \pi x^2 }$
1620648_1747568_ans_8503f253aefe47a79651900ad1c0215a.png

$H^{+} , He^{+}$ and

$O^{++}$ ions having same kinetic energy pass through a region of space filled with uniform magnetic field

$B$ directed perpendicular to the velocity of ions. The masses of the ions

$H^{+}, He^{+}$ and

$O^{++}$ are respectively in the ratio

$1 : 4 : 16$ . As a result

0%
$H^{+}$ ion s will be deflected most
0%
$O^{++}$ ions will be deflected least
0%
$He^{+}$ and $O^{++}$ ions will suffer same deflection
0%
All ions will duffer the same deflection

Explanation

The radius of the ion in magnetic field is:

$r=\dfrac{\sqrt{2mk}}{qB}$

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

$\Rightarrow r_{H}:r_{He}:r_{0}=\dfrac{\sqrt{1}}{1}:\dfrac{\sqrt{4}}{1}:\dfrac{\sqrt{16}}{2}=1:2:2$

Radius is smallest for

$H^{+}$ , so it is deflected most.

Two long parallel wires carry currents of

$20.0 \,A$ and

$10.0 \,A$ in opposite directions (Figure). Which of the following statements is true? More than one statement may be correct.

0%
In region I, the magnetic field is into the page and is never zero
0%
In region I, the magnetic field is into the page and is never zero.
0%
In region III, it is possible for the field to be zero.
0%
In region I, the magnetic field is out of the page and is never zero.
0%
There are no points where the field is zero.

Two long, parallel wires each carry the same current

$I$ in the same direction (Figure). Is the total magnetic field at the point

$P$ midway between the wires

0%
zero,
0%
directed into the page,
0%
directed out of the page,
0%
directed to the left, or
0%
directed to the right?

Explanation

The contribution made to the magnetic field at point

$P$ by the lower wire is directed out of the page, while the contribution due to the upper wire is directed into the page. Since point

$P$ is equidistant from the two wires, and the wires carry the same magnitude currents, these two oppositely directed contributions to the magnetic field have equal magnitudes and cancel each other.

Consider two parallel wires carrying currents in opposite directions in Figure. Due to the magnetic interaction between the wires, does the lower wire experience a magnetic force that is

0%
upward
0%
downward
0%
to the left,
0%
to the right, or
0%
into the paper?

Two parallel long wires carry currents

$18 \ A$ and

$3\ A$ . When the currents are in the same direction, the magnetic field at a point midway between the wire is

$B_1$ . If the direction of

$i_2$ is reversed, the field becomes

$B_2$ . Then the value of

$B_1/B_2$ is

0%
$5:7$
0%
$7:5$
0%
$3:5$
0%
$5:3$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 14 - 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 14 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

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