MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 3

A charged particle experiences magnetic force in the presence of magnetic field. Which of the following statement is correct?

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The particle is moving and magnetic field is perpendicular to the velocity.
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The particle is moving and magnetic field is parallel to velocity.
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The particle is stationary and magnetic field is perpendicular.
0%
The particle is stationary and magnetic field is parallel.

Quantity that is not affected by magnetic field is:

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Moving charge
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Change in magnetic flux
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Current flowing in conductor
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Stationary charge

State whether True or False :

Lenz's law defines the polarity of the induced voltage.

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True
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False

Identify the wrong statement.

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Current loop is equivalent to a magnetic dipole
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Magnetic dipole moment of a planar loop of area $A$ carrying current $l$ is $l^{2}A$
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Particles like proton, electron carry an intrinsic magnetic moment
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The current loop (magnetic moment $m$ ) placed in a uniform magnetic field, $B$ experiences a torque $\tau = m\times B$
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Ampere's circuit law is not independent of Biot Savart's law

Explanation

The second statement is wrong because magnetic dipole moment of planner loop of area

$A$ carrying current is

$IA$ . So, the option (b) is wrong.

A charged particle is moving along a magnetic field line. The magnetic force on the particle is

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Along its velocity
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Opposite to its velocity
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Perpendicular to its velocity
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Zero

Choose the correct alternative which matches second and third column with first column:

Column I	Column II	Column III
(I) Magnetic field is produced near current carrying conductor	(A) Right hand thumb rule	(a) Micheal Faraday
(II) Electric current is generated in a conductor moving in a magnetic field.	(B) Fleming's right hand rule	(b) Hans Oersted

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(I)-(B)-(a),(II)-(B)-(b)
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(I)-(A)-(b),(II)-(B)-(b)
0%
(I)-(B)-(b),(II)-(A)-(a)
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(I)-(A)-(b),(II)-(B)-(a)

A uniform magnetic field exists in the plane of paper pointing from left to right as shown in figure. In the field, an electron and a proton move as shown. The electron and the proton experiences:

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Forces both pointing into the plane of paper
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Forces both pointing out of the plane of paper
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Forces pointing into the plane of paper and out of the plane of paper respectively
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Forces pointing opposite and along the direction of the uniform magnetic field respectively.

A bar magnet of magnetic moment

$M$ is bent to form a semicircle. What is the magnetic moment of the bent magnet?

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

Explanation

Given: Magnetic moment=M

Solution: As we know the formula of magnetic moment

$M=m\times l$ ....(1)

Where l is the distance between two poles

According to the question bar magnet is converted into semicircle which means that length is converted into the circumference but remember there is no change in the pole strength(m)

So,

$l=\pi r$

$r=\frac{l}{\pi }$

$M^{'}=m\times 2r$ .......(2)

Where 2r= distance between two poles in semicircular form

Comparing eq (1) &(2) gives

$M^{'}=\frac{2M}{\pi }$

Hence the correct answer is B

An eleclron is travelling horizontallty towards east. A magnetic field in vertically downward (inside the plane of paper) direction exerts a force on the electron along

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East
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West
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North
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South

When a charged particle moves in a magnetic field its kinetic energy __________.

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Remains constant
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Increases
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Can decrease
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Become zero

A electron with kinetic energy

$k$ enters a region (i) uniform magnetic field (ii) non-uniform magnetic field and comes out with kinetic energy

$k_1$ and

$k_2$ respectively. What is relation between

$k_1, k_2$ and

$k$ ?

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$K_1 > K_2 > K$
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$K_1 = K_2 > K$
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$K_1 = K_2 = K$
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$K_1 < K_2 < K$

Explanation

Any type of magnetic field can not change kinetic energy of charged particle. So

$K_1 = K-2 = K$

The gyromagnetic ratio of an electron

$=$ __________ specific charge of an electron.

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$1$
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$2$
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$1/2$
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$4$

Explanation

Ratio

$=q/2m=1/2\times$ specific charge.

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{o}$ from the meridian is $n$ times the corresponding work done to turn it through an angle of $60^{o}$ from the meridian, where $n$ is given by :

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

Explanation

Work done in 1st case,

$W_1=MB(cos0^0-cos90^0)$

Work done in 2nd case,

$W_2=MB(cos0^0-cos60^0)$

$W_1=2W_2$

therefore,

$n=2.$

The distance between the wires of electric mains is 12cm. These wires exprience 4 mgwt per unit length. The value of current flowing in each wire will be if they carry current in same direction

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4.85A
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zero
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$4.85\times 10^{-2}A$
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$85\times 10^{-4}A$

Explanation

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

$4=\dfrac {2\times 10^{-7}\times i^2}{12\times 10^{-2}}$

$i=4\cdot 85A$

A magnet of moment $4 \ Am^{2}$ is suspended in a uniform magnetic field of induction $5\times 10^{-4} \ T$ with its axis at right angles. When the magnet is released from it's position, the kinetic energy that it gains in passing to the stable equilibrium position is :

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$10^{-4}$ J
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$10^{-3}$ J
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$2\times 10^{-3}J$
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$5\times 10^{-4}J$

Explanation

magnetic moment

$m=4\ Am^{2}$
magnetic field

$B=5\times 10^{-4}T$
Potential energy of dipole placed in uniform field

$\mu =-m.B$ ...(i)

From energy conservation principle:

$\Delta K+\Delta U=0$ ...(ii)

$K_{2}-K_{1}=(U_{1}-U_{2})$

$K_{1}=0$ ,

$U_{1}=0$

$U_{2}=-mB$ from equation (i)

$=-4\times 5\times 10^{-4}$

From equation (ii)

$K_{2}-0=0+20\times 10^{-4}$

$K_{2}=2\times 10^{-3} \ J$

The work done in rotating the magnet from the direction of uniform field to the opposite direction to the field is

$W$ . The work done in rotating the magnet from the field direction to half the maximum couple position is :

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$2\ W$
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$\dfrac{\sqrt{3}W}{2}$
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$\dfrac{W}{4}(2-\sqrt{3})$
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$\dfrac{W}{4}(1-\sqrt{3})$

Explanation

$\textbf{Hint}$ : Use the concept of dipole in a uniform magnetic field.

$\textbf{Step 1}$ :

Given the work done in rotating the magnet from the direction of uniform field to the opposite direction to the field is

$W$ .

Here

$\theta_1=0^{\circ}$ and

$\theta_2=180^{\circ}$

We know

$U=mBcos\theta$

$U_1=-mB cos0=-mB$

$U_2=-mBcos180^{\circ}=mB$

From the question we can write

$W=U_2-U_1$

$W=2mB$ ...........

$(1)$

Here at

$\theta=30^{\circ}$ Torque will be maximum when

$sin \theta=\dfrac{1}{2}$

We should find the work done in rotating the magnet from the field direction to the position of half the torque.

So,

$W_2=U_3-U_1$

$U_3=-mBcos30^{\circ}=mB\dfrac{\sqrt{3}}{2}$

$W_2=mB(1-\dfrac{\sqrt{3}}{2})$ ............

$(2)$

$\textbf{Step 2}$ :

From

$(1)$ and

$(2)$

$W_2=\dfrac{W}{2}(1-\dfrac{\sqrt{3}}{2})$

$W_2=\dfrac{W}{4}(2-\sqrt{3})$

Thus option C is correct.

A magnet of moment $1.2\ Am^{2}$ is kept suspended in a magnetic field of induction $2\times 10^{-6}$ . The work done in rotating it through 120 $^{o}$ is:

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$2.4\times 10^{-6}J$
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$4.8\times 10^{-6}J$
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$1.2\times 10^{-6}J$
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$3.6\times 10^{-6}J$

Explanation

$m=1.2Am^{2}$

$B=2\times 10^{-6}T$

$\theta =120^{o}$

$U=-mB\cos 0^o$

$U_{1}=-mB$

$=-2.4\times 10^{-6}J$

$U_{2}=-mB\cos 120^o$

$=\dfrac{mB}{2}$

$=1.2\times 10^{-6}J$

$w=U_{2}-U_{1}$

$=1.2\times 10^{-6}+2.4\times 10^{-6}$

$=3.6\times 10^{-6}$

An electron enters a magnetic field with a speed of

$10^{8} cm/s$ . The particle experiences a force due to the magnetic field and the speed of the electron

0%
will decrease
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will increase
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will remain constant
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may in crease or decrease

Explanation

A particle travelling perpendicular to a magnetic field would experience a force given by

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

This force, thus, acts perpendicular to the direction of motion of electron. Hence it only changes the direction of electron but not the speed of electron perpendicular to the direction of motion. This force would cause a constant centripetal acceleration. Thus the speed of electron does not change at all.

Assertion: When radius of a circular wire carrying current is doubled, its magnetic moment becomes four times

Reason: Magnetic moment is directly proportional to area of the loop

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Both A and R are true and R is the correct explanation of A.
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Both A and R are true and R is not correct explanation of A.
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A is true, but R is false
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A is false, but R is true

Explanation

$m$ (for circular wire)=

$I\pi r^{2}$ ( standard result for magnetic force on a current carrying wire)

$\dfrac{m_{1}}{r{_{1}}^{2}}=\dfrac{m_{2}}{r{_{2}}^{2}}$

$m_{2}=4m_{1}$

The magnetic moment of a bar magnet is 0.256 amp.m $^{2}$ . Its pole strength is 400 milli amp. m. It is cut into two equal pieces and these two pieces are arranged at right angles to each other with their unlike poles in contact (or like poles in contact). The resultant magnetic moment of the system is

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$\sqrt{2}\times 256\times 10^{-3}Am^{2}$
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$250\times 10^{-3}Am^{2}$
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$\dfrac{256}{\sqrt{2}}\times 10^{-3}Am^{2}$
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$\dfrac{128}{\sqrt{2}}\times 10^{-3}Am^{2}$

Explanation

Magnetic moment of bar magnet,

$M\propto l$

$M_1=\dfrac{M}{2}$

Therefore,

$M=\dfrac{0.256}{2}$

resultant magnetic moment=

$\sqrt{(\mu_1) ^2+(\mu_1) ^2}$

$\dfrac{0.256}{\sqrt{2}}=\dfrac{256}{\sqrt2} \times 10^{-3}$

Hence, Option c is answer.

An electron of mass m is accelerated through a potential difference of V and then it enters a magnetic field of induction B normal to the lines of force. Then the radius of the circular path is

0%
$\sqrt{\dfrac{2eV}{m}}$
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$\sqrt{\dfrac{2Vm}{eB^{2}}}$
0%
$\sqrt{\dfrac{2Vm}{eB}}$
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$\sqrt{\dfrac{2Vm}{e^{2}B}}$

Explanation

$KE=eV=\dfrac{1}{2}m_eV^2$
and

$R=\dfrac{m_eV}{eB}$

$=\dfrac{m_e}{eB}\times\sqrt{\dfrac{eV\times 2}{m_e}}$

$=\sqrt{\dfrac{2Vm_e}{eB^2}}$

A bar magnet of magnetic moment 3.0 A-m

$^{2}$ is free to rotate about a vertical axis passing through its centre. The magnet is released from rest from east - west position. Then the kinetic energy of the magnet as it takes North- South position is :
(horizontal component of earths magnetic field is 25

$\mu$ T)

0%
25 $\mu$ J
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75 $\mu$ J
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100 $\mu$ J
0%
12.5 $\mu$ J

Explanation

$m=3 Am^{2}$

$B=25\mu T$

$\Delta K.E.=\Delta P.E.$

$P=-m\cdot B$

$=-m B cos \theta$

$\theta _{1}=90$

$\theta _{2}=0$

Kinetic energy

$=P_{2}-P_{1}$

$=m B \cos \theta_{1} -m B \cos \theta_{2}$

$=3\times 25\mu$

$\therefore$ Kinetic energy of the Bar magnet will be 75

$\mu J$

A magnet is in stable equilibrium in a uniform magnetic field. It is deflected by $60^{o}$ and the workdone is equal to $W$ . In deflecting the magnet further by $30^{o}$ , Work done is:

0%
$\dfrac{W\sqrt{3}}{2}$
0%
$\dfrac{W}{4}$
0%
$\dfrac{W}{2}$
0%
$W$

Explanation

$U=m.B$

$U_{1}=-mB$

$U_{2}=-mB\cos60^o$

$=\dfrac{-mB}{2}$

$U_{3}=-mB\cos90^o$

$=0$

$W=U_{2}-U_{1}$

$=\dfrac{mB}{2}$

$W^{1}=U_{3}-U_{2}$

$=\dfrac{mB}{2}$

$W^{1}=W$

A magnet of magnetic moment

$M$ is rotated through 360

$^\circ$ in a magnetic field

$H$ . Work done will be

0%
$0$
0%
$2\ MH$
0%
$MH$
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$\pi MH$

Explanation

$\theta _{1}=0^{o}$

$\theta _{2}=360^{o}$

$U_{1}=-mBcos{0}$

$=-mB$

$U_{2}=-mBcos360$

$=-mB$

$w=U_{2}-U_{1}$

$=0$

When a charged particle moves through a magnetic field, the quantity which is not affected in the magnetic field is:

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particle velocity
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particle acceleration
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linear momentum of the particle
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kinetic energy of the particle

The force felt by an electron on entering into a magnetic field is independent of its

0%
Charge
0%
Strength of the field
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Mass
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Direction of its velocity

Explanation

$\vec{F_B} = q \vec{v} \times \vec{B}$
independent of mass

You are sitting in a room in which uniform magnetic field is present in vertically downward direction. When an electron is projected in horizontal direction, it will be moving in circular path with constant speed

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clockwise in vertical plane
0%
clockwise in horizontal plane
0%
anticlockwise in vertical plane
0%
anticlockwise in horizontal plane

Explanation

$F = q\vec{v} \times \vec{B}$
where

$q$ is

$-ve$
therefore, it will be moving in a circular path with constant speed clockwise in horizontal plane

The magnetic field

$\overline{dB}$ due to a small current element dl at a distance

$\vec{r}$ and carrying current ‘i’ is

0%
$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i\left ( \dfrac{\overline{dl}\times \bar{r}}{r} \right )$
0%
$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i^{2}\left ( \dfrac{\overline{dl}\times \bar{r}}{r^{2}} \right )$
0%
$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i^{2}\left ( \dfrac{\overline{dl}\times \bar{r}}{r} \right )$
0%
$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i\left ( \dfrac{\overline{dl}\times \bar{r}}{r^{3}} \right )$

For a given distance from a current element, the magnetic induction is maximum at an angle measured with respect to axis of the current. The angle is :

0%
$\dfrac{3\pi}{ 4}$
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$\dfrac{\pi }{4}$
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$\dfrac{\pi} {2}$
0%
$2\pi$

Explanation

Magnetic field due to small current element is given as

$B=\dfrac{\mu_o}{4\pi}\dfrac{\vec{idl}\times \hat{r}} {r^2}$

$B=\dfrac{\mu_o}{4\pi}\dfrac{idlr \sin\theta}{r^3}$

$\sin\theta=1$ for

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

An electron and a proton are injected into a uniform magnetic field at right angle to its direction with the same momentum. Then

0%
electrons path isless curved than protons path
0%
protons path will be less curved than electrons

path
0%
the paths of both will be equally curved
0%
both the trajectories will be straight

Explanation

R =

$\dfrac{mv}{QB} = \dfrac{P}{QB}$ [ P is same]
Q and B are same in magnitude in case of e and p their path will be equally curved

Magnetic field at a point on the line of current carrying conductor is

0%
maximum
0%
infinity
0%
zero
0%
finite value

Explanation

because angle between line and distance becomes zero.

$\therefore\overrightarrow{l}\times\widehat{r}$ becomes zero.

A cathode ray beam is bent into an arc of a circle of radius

$0.02 m$ by a field of magnetic induction

$4.55 mT$ . The velocity of electrons is:
(Given

$e=1.6\times 10^{-19}c$ and

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

0%
$2\times 10^{7}m/s$
0%
$3\times 10^{7}m/s$
0%
$1.6\times 10^{7}m/s$
0%
$3.2\times 10^{7}m/s$

Explanation

Radius in Magnetic field

$= \dfrac{mV}{eB}$

$\Rightarrow V = \dfrac{Ber}{m}$

$\Rightarrow V = \dfrac{4.55 \times 10^{-3} \times 1.6 \times 10^{-19} \times 0.02}{9.1 \times 10^{-31}}$

$\Rightarrow V = 1.6 \times 10^{7}m/s$

The magnetic field due to a current element is independent of :

0%
current through it
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distance from it
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its length
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nature of meterial

Explanation

$B=\dfrac{\mu_o}{4\pi}\dfrac{i\overrightarrow{dl}\times\overrightarrow{r}}{r^3}$

So it depends on all three : current , distance and length
but not on nature of material

$\beta -$ particle enters a magnetic field making an angle of

$45^{o}$ with the field lines. The path of the particle is

0%
circular
0%
elliptical
0%
spiral
0%
a straight line

Explanation

the

$v_x = \dfrac{v}{\sqrt 2}$ will be responsible for circular motion and

$v_y$ will take it in

$+Y$ direction thus making it a spiral path

An electron and a proton enter a magnetic field with equal velocities. The particle that experiences more force is

0%
electron
0%
proton
0%
both experience same force
0%
it cannot be predicted.

Explanation

as charges and velocities are same

$\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$
will be same, but in opposit direction

A magnetic field exerts no force on

0%
a stream of electrons
0%
a stream on protons
0%
unmagnetised piece of iron
0%
stationary charge.

Explanation

$F=q(\overrightarrow{v}\times \overrightarrow{B})$
if

$v=0\rightarrow F=0$

A charged particle enters into a uniform magnetic field the parameter that remains constant is

0%
velocity
0%
momentum
0%
kinetic energy
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angular velocity

A charged particle is moving with velocity

$v$ in a magnetic field of induction

$B$ . The force on the particle will be maximum when

0%
$v$ and $B$ are in the same direction
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$v$ and $B$ are in Opposite direction
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$v$ and $B$ are perpendicular
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$v$ and $B$ are at an angle of $45^{o}$

Explanation

Force

$= q(\vec{V}\times \vec{B})$

$= q VB \sin 0$
Where,

$0$ is angle between velocity and magnetic field.
So,

$\sin 0$ is maximum when

$0$ is

$90^o$ or velocity is perpendicular to magnetic field.

A charge moving with velocity

$v$ in

$X-$ direction is subjected to a field of magnetic induction in the negative

$X$ direction. As a result the charge will :

0%
remain unaffected
0%
start moving in a circular path in $Y-Z$ plane
0%
retard along $X-$ axis
0%
move along a helical path around $X -$ axis

Explanation

velocity

$V$ is in

$X$ direction and magnetic field

$B$ is in negative

$X$ direction hence angle between the velocity and magnetic field is

$180^o$ .

In given case force on moving charge due to magnetic field is given by,

$F=qVBsin180^o=0$

Hence charge remains unaffected.

The mono energetic beams of electrons moving along +y direction enter a region of uniform electric and magnetic fields. If the beam goes straight through these simultaneously, then field B and E are directed possibly along

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

Explanation

$\overrightarrow{V} =V({\widehat{j}})$
Hence , B and E should be perpendicular to each other and also perpendicular to the velocity . Hence , B and E should be in z - axis and x axis
If B is towards + z axis , Force = q(V \times B ) is in + x direction . To balance this , E has to be in -x direction

A free charged particle moves through a magnetic field. The particle may undergo a change in

0%
speed
0%
energy
0%
direction of motion
0%
magnitude of the velocity

Explanation

Hint:

When any charge particle with charge $q$ is moving with velocity $v$ in magnetic field $B$ . A force acts on it which is given by,

$F = qvB sin\theta$

where, $\theta$ is the angle between direction of velocity of charge and Magnetic Field.

Correct Option is C.

Explanation for correct answer:

$\bullet$ Whenever any charge

$q$ is moving with velocity

$v$ in an magnetic field

$B$ , a force acts on it, which is given by

$F = qvB sin\theta$

or,

$F=q(v \times B)$

Note that, the force is perpendicular to

$v$ and

$B$ .

$\bullet$ The force is perpendicular to the direction of motion of the charge. Thus it do not accelerate the charge, in other words it do not changes the magnitude of velocity of charge, it just changes the direction of motion of the charge.

A charged particle moving in a magnetic field experiences a resultant force

0%
in the direction opposite to that of the field.
0%
in the direction opposite to that of its velocity
0%
in the direction perpendicular to both field &

its velocity
0%
in the direction parallel to the field

Explanation

$\overrightarrow{F}=q(\overrightarrow{V}\times\overrightarrow{B})$
F is perpendicular to both velocity and field.

A proton and an electron enter a region with equal speed in which a magnetic field is suddenly switched on. The force experienced by them are

0%
equal and opposite
0%
different in magnitude but same in direction
0%
in the ratio of 1840
0%
same in magnitude and direction

Explanation

charge is equal in magnitude but opposite in sign

$\therefore$ forces are equal in magnitude but opposite in direction

When a charged particle moves in a uniform magnetic field

0%
it gains energy from the field
0%
it loses energy to the field
0%
it neither gains nor changes energy and momentum
0%
momentum changes but not energy

A proton enters a magnetic field with a velocity of

$2.5\times 10^{7}ms^{-1}$ making an angle

$30^{0}$ with the magnetic field. The force on the proton is

$(B=25T)$

0%
$1.25\times 10^{-11}N$
0%
$2.5\times 10^{-11}N$
0%
$5.0\times 10^{-11}N$
0%
$7.5\times 10^{-11}N$

Explanation

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

$F-qvB sin\theta$

$F=1\cdot 6\times 10^{-19}\times 2\cdot 5\times 10^7\times 25\times sin 30$

$F=5\times 10^{-11} N$

A particle of mass

$M$ and charge

$Q$ moving with velocity

$\vec{v}$ describes a circular path of radius

$R$ when subjected to a uniform transverse magnetic field of induction

$\vec{B}$ . The work done by the field when the particle completes one full circle is

0%
$MV^{2}R^2$
0%
$0$
0%
$VBQR$
0%
$2\pi VBQR$

Explanation

$\bar F=q(\bar {V}\times \bar{B})$
Here

$\bar F$ is perpendicular to the velocity of particle.
So

$P=\bar {F}\cdot \bar {V}$

$=0$
So the workdone by the particle

$=0$

A current of one ampere is passed through a straight wire of length 2 metre. The magnetic field at a point in air at a distance of 3 m from one end of the wire but lying on the axis of the wire will be:

0%
$\mu _{0}/2\pi$
0%
$\mu _{0}/4\pi$
0%
$\mu _{0}/8\pi$
0%
Zero

Explanation

Hint:

The Magnetic Field at a Distance $a$ from a long wire carrying current $I$ is given by,

$B=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{I}{a}\left( \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right)$

Where angles ${{\phi }_{1}}$ and ${{\phi }_{2}}$ are as shown in the figure.

Correct Option is D.

Explanation for the correct answer:

Step 1: Note all the given values.

The magnetic field produced by a current-carrying straing conductor carrying current $I$ at a point P at a distance $r$ from the wire is calculated as follows:

$B=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{I}{a}\left( \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right)$

In the given problem, we must calculate the Magnetic Field at a distance of $3m$ from one end of the wire, which is lying on the wire's axis.

Therefore, ${{\phi }_{1}}=0\,and\,{{\phi }_{2}}=180$

Step 2: Calculate Field at given point.

Magnetic Field at a point at distance $3m$ from one end of the wire lying on the axis of the wire is,

$B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{I}{a}\left( {\sin 0 + \sin 180} \right)$

$\Rightarrow B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{I}{a}\left( {0 + 0} \right)$

$\Rightarrow B = 0$

As a result, the net Magnetic Field at a position $3m$ away from one end of the wire lying on the wire's axis is zero.

Singly ionized helium(x), ionized deuteron(y), alpha(z) particles are projected into a uniform magnetic field

$3 \times 10^{-4}$ Tesla with velocities

$10^{5} ms^{-1}$ ,

$0.4 \times 10^{4} ms^{-1}$ and

$2 \times 10^{3} ms^{-1}$ respectively. The correct relation between the ratio of the angular momentum to the magnetic moment of the particles is :

0%
$x>y= z$
0%
$x < y < z$
0%
$x< z < y$
0%
$z > x > y$

Explanation

The magnetic moment of the particle is given by

$M = \dfrac{qL}{2m}$

thus rearranging

$\dfrac{L}{M} = \dfrac{2m}{q}$

So,

$\dfrac{L}{M}\ \alpha \ \dfrac{m}{q}$

For deuteron and alpha particles

$\dfrac{m}{q}$ is same and for singly ionized helium

$\dfrac{m}{q}$ is more than alpha particles. So, correct order is

$x > y = z$

A proton is rotating along a circular path with kinetic energy K in a uniform magnetic field B.If the magnetic field is made four times, the kinetic energy of rotation of proton is

0%
16K
0%
8K
0%
4K
0%
K

Explanation

The force acting on particle moving with speed

$v$ is given by

$F=qvB$

This provides the centripetal acceleration to the particle.

hence

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

$\implies v=\dfrac{qBr}{m}$

Hence the kinetic energy=

$K=\dfrac{1}{2}mv^2\propto B^2$

Hence when magnetic field becomes 4 times, the kinetic energy of proton becomes 16 times.

A particle of mass

$0.6$

$g$ and having a charge of

$25$

$nC$ is moving horizontally with a uniform velocity

$1.2 \times10^{4}$

$m/s$ in a uniform magnetic field, then the value of the magnetic induction is (Take

$g=10 \ m/s^2$ )

0%
$zero$
0%
$10 T$
0%
$20 T$
0%
$200 T$

Explanation

Particle is moving with uniform velocity therefore no force is acting on the particle.
So

$mg- qvB=0$

$0\cdot 6\times 10^{-3}\times 10=25\times 10^{-9}\times 1\cdot 2\times 10^4\times B$

$B=20 T$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 3 - 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 3 - MCQExams.com

0:0:4

Practice Class 12 Medical Physics Quiz Questions and Answers

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