Assertion is correct but Reason is incorrect

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Both Assertion and Reason are incorrect

antiparallel current attract each other.

equal magnitudes of antiparallel current attract each other.

MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 1

A particle with a specific charge s is fired with a speed v towards a wall at a distance d, perpendicular to the wall. What minimum magnetic field must exist in this region for the particle not to hit the wall?

0%
$\frac{v}{sd}$
0%
$\frac{2v}{sd}$
0%
$\frac{v}{2sd}$
0%
$\frac{v}{4sd}$

The distance between the supply wires of an electric-mains is

$12cm$ . These wires experience

$4mg$ weight per unit length. Current flowing in both the wire will be if they carry current in same direction.

0%
zero
0%
$4.85A$
0%
$4.85A$
0%
$4.85\times {10}^{-4}A$

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii

$r_e, r_p, r_{\alpha}$ respectively in a uniform magnetic field B. The relation between

$r_e, r_p, r_{\alpha}$ is?

0%
$r_e < r_p < r_{\alpha}$
0%
$r_e < r_{\alpha} < r_p$
0%
$r_e > r_p = r_{\alpha}$
0%
$r_e < r_p = r_{\alpha}$

Explanation

We know that

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

Also

$mv = \sqrt{2Km}$

where K is the kinetic energy

Hence

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

Since

$K$ and

$B$ are constants,

$r = K' \sqrt{m}{B}$

Since

$m_{\alpha}= 4 m_{p}$

and

$q_{\alpha}= 2 q_{p}$ ,

$r_{\alpha}= r_{p}$

and Since

$m_{p}> m_{e}$

and

$q_{p}= q_{e}$

$r_{p}>r_{e}$

A particle of charge

$-16\times 10^{-18}$ coulomb moving with velocity 10

$\mathrm{m}\mathrm{s}^{-1}$ along the x-axis enters a region where a magnetic field of induction

$\mathrm{B}$ is along the

$\mathrm{y}$ -axis, and an electric field of induction

$\mathrm{B}$ is along the

$\mathrm{y}$ -axis, and an electric field of magnitude

$10^{4}\mathrm{V}/\mathrm{m}$ is along the negative

$\mathrm{z}$ -axis. lf the charged particle continues moving along the

$\mathrm{x}$ -axis, the magnitude of

$\mathrm{B}$ is

0%
$10^{3}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$
0%
$10^{5}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$
0%
$10^{16}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$
0%
$10^{-3}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$

Explanation

Particle travels along

$x$ axis

Hence

${ V }_{ y }={ V }_{ z }=0$

Electric feild is along the negative

$z$ axis

${ E }_{ x }={ E }_{ y }=0$

Therefore net force on the particle

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

Resolve the motion along three coordinate axis

${ a }_{ x }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ x }+{ V }_{ y }{ B }_{ z }-{ V }_{ x }{ B }_{ y } \right) \\ { a }_{ y }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ y }+{ V }_{ z }{ B }_{ x }-{ V }_{ xz } \right) \\ { a }_{ z }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ z }+{ V }_{ x }{ B }_{ y }-{ V }_{ y }{ B }_{ x } \right)$

Since,

${ V }_{ y }={ V }_{ z }=0\\ { E }_{ x }={ E }_{ y }=0\\ { B }_{ x }={ B }_{ z }=0$

${ a }_{ x }={ a }_{ y }=0\\ { a }_{ z }=\cfrac { q }{ m } \left( -{ E }_{ z }+{ V }_{ x }{ B }_{ y } \right)$

Again

${ a }_{ z }=0$ , as the particle transverse through the region undeflected

${ E }_{ z }={ V }_{ x }{ B }_{ y }$

${ B }_{ y }=\cfrac { { E }_{ z } }{ { V }_{ x } } =\cfrac { { 10 }^{ 4 } }{ 10 } ={ 10 }^{ 3 }wb/m^{ 2 }$

Which of the following expressions are applicable to the moving coil galvanometer?

0%
$\displaystyle \overrightarrow{F_m}=q(\overrightarrow {V}\times\overrightarrow{B})$
0%
$\displaystyle B=B_0\tan\theta$
0%
$\displaystyle \overrightarrow {\tau}=\overrightarrow{M}\times\overrightarrow{B}$
0%
none of these

Explanation

Torque

$\overrightarrow{\tau} = \overrightarrow{M} \times \overrightarrow{B}$

Torque is applicable to the moving coil galvanometer.

A charged particle moves through a magnetic field perpendicular to its direction. Then

0%
the momentum changes but the kinetic energy is constant
0%
both momentum and kinetic energy of the particle are not constant
0%
both, momentum and kinetic energy of the particle are constant
0%
kinetic energy changes but the momentum is constant

Explanation

The work done is the change in kinetic energy (KE). Since the work done by the magnetic field is zero. So, KE

$=$ constant.

But the particle is moving through a magnetic so velocity changes it means momentum also changes.

Two long conductors, separated by a distance

$d$ carry currents

$I_1$ and

$I_2$ in the same direction. They exert a force

$F$ on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to

$3d$ . The new value of the force between them is:

0%
$-2F$
0%
$\dfrac{F}{3}$
0%
$\dfrac{-2F}{3}$
0%
$\dfrac{-F}{3}$

Explanation

$\text{Magnetic field due to a wire at a distance}\ d\ \text{is}\ B=\dfrac{\mu_0 I_1}{2\pi d}$

$\text{Force on other wire due to this magnetic field of length} \ l \ \text{is}\ F=BI_2l=\dfrac{\mu_0 I_1}{2\pi d}I_2 l$

$\text{Therefore force per unit length}=F/l=\dfrac{\mu_0 I_1I_2}{2\pi d}$

$\Rightarrow F/l\propto \dfrac{I_1I_2}{d}$

$\text{When current becomes two times and distance is increased to three times,}$

$F_1=\dfrac{2F}{3}$

$\text{Since the direction of current is also reversed in one of them, the direction of force also reverses.}$

$\Rightarrow F^{'}=\dfrac{-2F}{3}$

Two long straight parallel wires, carrying (adjustable) current

$I_1$ and

$I_2$ , are kept at a distance d apart. If the force 'F' between the two wires is taken as 'positive' when the wires repel each other and 'negative' when the wires attract each other, the graph showing the dependence of 'F', on the product

$I_1I_2$ , would be

Explanation

The force between two wires per unit length is:

$F=\dfrac{\mu _0}{2 \pi d} I_1I_2$

When,

$I_1$ and

$I_2$ are in in the same direction, the wires attract each other, and

$F \lt 0$
When,

$I_1$ and

$I_2$ are in opposite direction, the wires repel each other, and

$F \gt 0$
The plot is a straight line since

$F \propto I_1I_2$
The slop is negative since sign of force is opposite to the product

$I_1I_2$ ( same direction product is positive, opposite direction product is negative.

A particle of mass M and charge Q moving with velocity

$\vec v$ describe a circular path of radius R when subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes one full circle is

0%
$\displaystyle \left ( \frac{Mv^2}{R} \right ) 2 \pi R$
0%
$zero$
0%
$BQ2 \pi R$
0%
$BQv2 \pi R$

A particle of mass

$\mathrm{M}$ and charge

$\mathrm{Q}$ moving with velocity

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

$\mathrm{R}$ when subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes one full circle is

0%
$(\displaystyle \frac{\mathrm{M}\mathrm{v}^{2}}{\mathrm{R}})2\pi \mathrm{R}$
0%
zero
0%
$\mathrm{B}2\pi \mathrm{R}$
0%
$\mathrm{B}\mathrm{v}2\pi \mathrm{R}$

A rectangular loop of sides 10 cm and 5 cm carrying a current I and 12 A is placed in different orientations as shown in the figures above.
If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?

0%
(A) and (B), respectively
0%
(A) and (C), respectively
0%
(B) and (D), respectively
0%
(B) and (C), respectively

Explanation

$U=-\vec { M. } \vec { B } =-MBCos\theta$
So,

$U_{min}$ when

$\theta=0^0 \Rightarrow$ Stable equilibrium

$U_{max}$ when

$\theta=180^0 \Rightarrow$ Untable equilibrium

A loop carrying current

$I$ lies in the

$x$ -

$y$ plane as shown in the figure. The unit vector

$\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is

0%
$a^{2}I\hat{k}$
0%
$(\displaystyle \frac{\pi}{2}+1)a^{2}I\hat{k}$
0%
$-(\displaystyle \frac{\pi}{2}+1 )a^{2}I\hat{k}$
0%
$(2\pi+1)a^{2}I\hat{k}$

Explanation

Magnetic moment

$(M)=I.A = I \times (2 \times \dfrac{\pi}{4} a^2 + a^2) = (\dfrac{\pi}{2} + 1)a^{2} I\hat k$

An electron traveling with a speed

$\mathrm{u}$ along the positive

$\mathrm{x}$ -axis enters into a region of magnetic field where

$\mathrm{B}=-\mathrm{B}_{0}\hat{\mathrm{k}}(\mathrm{x}>0)$ . It comes out of the region with speed

$\mathrm{v}$ then :

0%
v = u at y > 0
0%
v = u at y < 0
0%
v > u at y > 0
0%
v > u at y < 0

Explanation

Charged particle moves in a circular trajectory with constant speed

$u$ , since force always acts perpendicular to the motion. It is not able to change its speed. It just tends to come out after one circle is completed with speed

$u$ .

Two identical bar magnets are fixed with their centres at a distance d apart. A stationary charge Q is placed at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure.
The force on the charge Q is

0%
$zero$
0%
directed along OP
0%
directed along PO
0%
directed perpendicular to the plane of paper

A proton and alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:

0%
1 Me V
0%
4 Me V
0%
0.5 MeV
0%
1.5 MeV

Explanation

The radius of circular path is

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

So,

$\dfrac{r_{\alpha}}{r_p}=\sqrt{\dfrac{m_{\alpha}E_{\alpha}}{m_pE_p}}\dfrac{q_p}{q_{\alpha}}$

or,

$1=\sqrt{\dfrac{E_{\alpha}}{E_p}}$ as radius of both particles are equal and the mass of alpha particle is 4 times of proton and charge is 2 times of proton.

Thus,

$E_{\alpha}=E_p=1$ MeV

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. the energy required to rotate it by

$60^o$ is

$W$ , Now the torque required to keep the magnet in this new position is.

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

Explanation

Hint:-Use potential energy formula to calculate the energy required to rotate the magnet and hence calculate torque.

Step 1: Calculate energy required to rotate magnet by $60^{\circ}$ .

Energy required or work done in moving from

$\theta=0^{\circ}$ to

$\theta=60^{\circ}$ is given by

$W=-mBcos 60^{\circ}-(- mBcos 0)=\dfrac{1}{2}mB$

Step 2: Calculate the torque required to keep the magnet in the new position.

The torque acting on a bar magnet kept in a magnetic field

$\vec{B}$ is given by

$\tau=\vec{m}\times \vec{B}=mBsin\theta$

$\tau=mBsin60^{\circ}$

$\tau=\dfrac{\sqrt3}{2}mB=\sqrt3W$

$\textbf{Hence option C correct}$

A short bar magnet of magnetic moment

$0.4 J {T}^{-1}$ is place in a uniform magnetic field of

$0.16 T$ . The magnet is stable equilibrium when the potential energy is

0%
$-0.064 J$
0%
Zero
0%
$-0.082 J$
0%
$0.064$

Explanation

For stable equilibrium

$U = -MB$

$= -\left(0.4\right) \left(0.16\right)$

$= -0.064 J$ .

Current sensitivity of a moving coil galvanometer is

$5$ div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is

$20$ div/V. The resistance of the galvanometer is?

0%
$250$ $\Omega$
0%
$40$ $\Omega$
0%
$500$ $\Omega$
0%
$25$ $\Omega$

Explanation

Current sensitivity

$I_g=\dfrac{NBA}{K}=\dfrac{5div}{mA}=5000\displaystyle\frac{div}{A}$

Voltage sensitivity $v_s=\dfrac{NBA}{KR}=20\dfrac{div}{v}$

$\Rightarrow\displaystyle\frac{I_g}{v_s}=R$

$\Rightarrow R=\displaystyle\frac{5000}{20}=250\Omega$

Option 1 is correct.

A proton carrying

$1MeV$ kinetic energy is moving in a circular path of radius

$R$ in uniform magnetic field. What should be the energy of an

$\alpha$ -particle to describe a circle of same radius in the same field?

0%
$1MeV$
0%
$0.5MeV$
0%
$4MeV$
0%
$2MeV$

Explanation

For a particle of mass m and charge Q in a magnetic field, the radius of the path is given by,

$r=\cfrac{\sqrt 2m(KE)}{qB}$

$q\propto \sqrt {m(KE)}$

$\cfrac{e}{2e}=\sqrt {\cfrac{({m}_{p})(IMeV)}{(4{m}_{p})(KE)}}$

$\cfrac{1}{4}=\cfrac{1}{4(KE)}$

$KE=1MeV$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

An uncharged particle is moving with a velocity of

$\vec { v }$ in non-uniform magnetic field as shown. Velocity

$\vec { v }$ would be:

0%
maximum at $A$ and $B$
0%
minimum at $A$ and $B$
0%
maximum at $M$
0%
same at all points

Explanation

Magnetic force

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

The particle being uncharged,

$q=0$ .

So, magnetic force on the particle,

$\vec{F_m}=\vec{0}$

Since there is no external force acting on the particle, its velocity

$\vec{v}$ will be same at all points.

Two wires carrying

0%
parallel current repel each other.
0%
antiparallel current attract each other.
0%
antiparallel current repel each other.
0%
equal magnitudes of antiparallel current attract each other.

Explanation

The direction of magnetic field can be determined from right hand thumb rule. The direction of magnetic force is given by the cross product

$\vec { F } =i(\vec { l } \times \vec { B } )$ .

In the case of two wires carrying parallel current, this force on one wire due to the magnetic field of the other is directed towards the other wire. So, wires carrying parallel current attract each other.

In the case of two wires carrying anti-parallel current, this force on one wire due to the magnetic field of the other is directed away from the other wire. So, wires carrying anti-parallel current repel each other.

A current carrying small loop behaves like a small magnet. If

$A$ be its area and

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

0%
$M/A$
0%
$A/M$
0%
$MA$
0%
$Am^{2}$

Explanation

Magnetic moment is defined as

$M = iA$

$\Rightarrow i = \dfrac {M}{A}$ .

When a charged particle is acted on only by a magnetic force, its:

0%
potential energy changes
0%
its kinetic energy changes
0%
total energy changes
0%
energy does not change

Explanation

Magnetic force acting on a charge

$q$ travelling with velocity

$v$ is given by

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

Since the force is perpendicular to the velocity, work done by force on the object is always zero.

Hence the total energy of the charged particle does not change.

Hence correct answer is option D.

When

$0.50\overset {\circ}{A}$ X-rays strike a material, the photoelectrons from the

$k$ shell are observed to move in a circle of radius

$23\ mm$ in a magnetic field of

$2\times 10^{-2}$ tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of

$k-shell$ electrons?

0%
$3.5\ keV$
0%
$6.2\ keV$
0%
$2.9\ keV$
0%
$5.5\ keV$

Explanation

As we know,

$F = qvB = m \dfrac {v^{2}}{R} \Rightarrow v = \dfrac {q}{m}BR$
The kinetic energy of the photoelectron

$\dfrac {1}{2} mv^{2} = \dfrac {1}{2} \dfrac {e^{2}B^{2}R^{2}}{m}$

$= \dfrac {1}{2} \dfrac {(1.6\times 10^{-19})^{2} (2\times 10^{-2})^{2} (23\times 10^{-3})^{2}}{(9.1\times 10^{-31})}$

$= 2.97\times 10^{-15}J$

$= \dfrac {2.97\times 10^{-15}}{1.6\times 10^{-19}} = 18.36\ keV$
Energy of the incident photon

$= \dfrac {hc}{\lambda}$

$= \dfrac {12.4}{0.50} = 24.8\ keV$
Therefore, Binding energy

$= 24.8 - 18.6 = 6.2\ keV$ .

Magnetic moment of bar magnet is

$M$ . The work done in turning the magnet by

$90^o$ in direction of magnetic field

$B$ will be

0%
Zero
0%
$\displaystyle\frac{1}{2}$ $MB$
0%
$3$ $MB$
0%
$MB$

Explanation

Potential energy of a magnetic moment at an angle

$\theta$ with the magnetic field =

$-MB\cos\theta$

Thus work done in rotating from angle

$\theta_1$ to

$\theta_2=MB(cos\theta_1-cos\theta_2)$

$=MB(cos0^{\circ}-cos90^{\circ})$

$=MB(1-0)$

$=MB$

The correct figure showing Fleming's Left Hand Rule is :

Two like poles of strength

$m_{1}$ and

$m_{2}$ are far distance apart. The energy required to bring them

$r_{0}$ distance apart is

0%
$\displaystyle \dfrac{\mu _{0}}{4\pi }\frac{m_{1}m_{2}}{r_{0}}$
0%
$\displaystyle \dfrac{\mu _{0}}{8\pi }\frac{m_{1}m_{2}}{r_{0}}$
0%
$0$
0%
$\displaystyle \dfrac{\mu _{0}}{16\pi }\frac{m_{1}m_{2}}{r_{0}^2}$

Explanation

$V_{1}=\dfrac{\mu _{0}m_{1}}{4\pi \Omega _{0}}$

$V_{1}=$ magnetic potential

$m_{1}=$ pole strength

$\Omega _{o}$ =distance from

$m_1$

$Energy=Vm =\dfrac{\mu _{0}m_{1}m_{2}}{4\pi \Omega _{0}}$

The work required to rotate a magnetic needle by 60 $^{0}$ from equilibrium position in a uniform magnetic field is W. The torque required to hold it in that position is

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

Explanation

$\theta _{1}=0$

$\theta _{2}=60$

$U_{1}=-mBcos0$

$=-mB$

$U_{2}=-mBcos60^{0}$

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

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

$=\dfrac{mB}{z}$

$z=mBsin\theta$

$=mBsin60$

$=\dfrac{mB}{z}\sqrt{3}$

$z=w\sqrt{3}$

A magnet of moment 4Am $^{2}$ is kept suspended in a magnetic field of induction $5\times 10^{-5}T$ . The workdone in rotating it through 180 $^{0}$ is

0%
$4\times 10^{-4}J$
0%
$5\times 10^{-4}J$
0%
$2\times 10^{-4}J$
0%
$10^{-4}J$

Explanation

Torque acting on a magnetic dipole of moment

$M$ kept in uniform magnetic field=

$MBsin\theta$

Therefore work done in rotating a magnet of moment

$M$ by an angle

$\theta=MB(1-\cos\theta)=MB(1-(-1))=2MB$

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

Assertion (A): A magnet remains stable, If it aligns itself with the field

Reason (R): The P.E. of a bar magnet is minimum, if it is parallel to magnetic field.

0%
Both A and R are true and R is the correct explanation of A.
0%
Both A and R are true and R is not correct explanation of A.
0%
A is true, But R is false
0%
A is false, But R is true

Explanation

The magnet remains stable,if it align itself with the field as its potential will be minimum in that position.

$\mu=-\vec{M}.\vec{B}$
=

$-MB\cos \Theta$

The potential energy will be minimum if

$\Theta =0$
i.e.

$-MB$

$\therefore$ If it is parallel to magnet field

A bar magnet of magnetic moment

$2Am^{2}$ is free to rotate about a vertical axis passing through its center. The magnet is released from rest from east - west position. Then the KE of the magnet as it takes N-S position is

$(B_{H}=25\mu T)$

0%
$25 \mu J$
0%
$50\mu J$
0%
$100\mu J$
0%
$12.5\mu J$

Explanation

$m=2Am^{2}$

$B_{H}=25\mu T$

$\theta _{1}=\pi /2$

$\theta _{2}=0$

$\mu _{1}=-mB\cos\pi /2$

$=0$

$\mu _{2}=-mB\cos0^o$

$=-2\times 25\mu T$

$k_{1}=0$

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

$k_{2}=50\mu J$

A vertical straight conductor carries a current vertically upwards. A point

$P$ lies to the east of it at a small distance and another point

$Q$ lies to the west at the same distance the magnetic field at

$P$ is :

0%
greater than at $Q$
0%
same as at $Q$
0%
less than at $Q$
0%
greater or less than at $Q$ depending upon the strength of current

The ratio of magnetic potential due to magnetic dipole at the end to that at the broad side on a position for the same distance from it is :

0%
zero
0%
$\infty$
0%
1
0%
2

Explanation

Magnetic potential(V) at any point due to bar magnet is given by

$V = \dfrac{\mu_0 M \cos \theta}{4 \pi r^2}$
where, variables have their usual meanings.

Magnetic potential due to magnetic dipole along axial line i.e. end on position,

$\theta = 0 \Rightarrow \cos \theta = 1$

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

And magnetic potential due to magnetic dipole along equatorial line i.e. on broad side,

$\theta = 90^\circ \Rightarrow \cos \theta = 0$

$\therefore V_e= 0$

Hence the ratio,

$\dfrac {Va}{Ve} = \infty$

Maximum P.E. of magnet of moment M situated in a magnetic field of induction B, is

0%
$\dfrac {1}{2}MB$
0%
$\dfrac {M}{B}$
0%
$2MB$
0%
$MB$

Explanation

$PB=-MB\cos\theta$

$\text{PE is max if}$

$\theta = \pi$

$\therefore PB_{max}=MB$

A short bar magnet placed with its axis at

$30^{o}$ with a uniform external magnetic field of

$0.16\ T$ experiences a torque of magnitude

$0.032 Nm$ If the bar magnet is free to rotate, its potential energies when it is in stable and unstable equilibrium are respectively

0%
$-0.064\ J, +0.064\ J$
0%
$-0.032\ J, +0.032\ J$
0%
$+0.064\ J, -0.128\ J$
0%
$0.032\ J, -0.032\ J$

Explanation

$\theta =30^{o}$

$B=0016\ T$

$z=0.032$

$z=mB\sin\theta$

$0.032=m\times \dfrac{1}{2}\times 0.16$

$m=0.4\ Am$

$U=-\vec{m}.\vec{B}$

$U_{max}=mB = 0.4\times 0.16=0.064\ T$

$U_{min}=-mB = -0.4\times 0.16= -0.064\ T$

The dimensional formula for magnetic moment is :

0%
$[M^{0}L^{1}T^{0}A^{2}]$
0%
$[M^{0}L^{2}T^{0}A^{1}]$
0%
$[M^{0}L^{2}T^{0}A^{2}]$
0%
$[M^{0}L^{0}T^{1}A^{1}]$

Explanation

While solving questions related to finding dimensional formulae, think about writing the given physical quantity in terms of other physical quantities whose dimensional formula is known.

The magnitude of magnetic moment for a closed loop is proportional to:

$Current[M^0L^0T^0A^1]$
$\text{Area of the loop}[M^0L^2T^0A^0]$

Therefore the dimensional formula for magnetic moment =

$[M^0L^2T^0A^1]$

A charge q is moving with a velocity v parallel to a magnetic field B. Force on the charge due to magnetic field is

0%
qvB
0%
qB/v
0%
Zero
0%
Bv/q

Explanation

$F + q(\vec{V}\times \vec{B})$

$\overrightarrow{V}||\overrightarrow{B},\theta=0$

$F=0$

The following cases in which no force is exerted by a magnetic field on a charge is

0%
moving with constant velocity
0%
moving in a circle
0%
at rest
0%
moving along a curved path

Explanation

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

$v=0, F_B=0$

The force acting on a charge q moving with a velocity V in a magnetic field of induction B is given by :

0%
$\dfrac{q}{\vec{V}\times \vec{B}}$
0%
$\dfrac{\vec{V}\times \vec{B}}{q}$
0%
$q(\vec{V}\times \vec{B})$
0%
$(\vec{V}.\vec{B})q$

A proton moving in a straight line enters a strong magnetic field along the field direction then its path and velocity will be

0%
path is circular but speed constant
0%
path is same but velocity increases
0%
path is same and velocity remains constant
0%
path is same and motion is retarded

Explanation

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

$= q VB sin 0$

$= q VB sin 0^0$

$(\because \text{V and B has same direction})$

$= 0$

$(\because sin0^0 = 0)$
So, there is no force, so, path is same and velocity is constant.

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%
retard along X-axis
0%
move along a helical path around X axis
0%
remain unaffected
0%
starts moving in a circular path

Explanation

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

$= q VB sin 0$

$= q VB sin(180^0)$

$(\because\text { velocity is in X-direction and B is in -X-direction})$

$= 0$

$(\because sin180^0 = 0)$
So, charge will be more unaffected.

A current loop placed in a magnetic field behaves like a

0%
magnetic dipole
0%
magnetic substance
0%
magnetic pole
0%
non magnetic substance

Explanation

Current loop placed in a magnetic field behaves like a magnetic dipole of magnetic moment

$M = NiA$

The dipole moment of a current loop is independent of

0%
current in the loop
0%
number of turns
0%
area of the loop
0%
magnetic field in which it is situated

Explanation

Dipole moment of a current loop

$M = N\times i\times A$
So, it is dependent on current in the loop, number of turns and area of the loop.
It is independent of magnetic field in which it is situated.

If an electron of velocity is

$2i+4j$ is subjected to magnetic field of

$4k$ , then, for the electron ,

0%
speed will change
0%
path will change
0%
velocity is Constant
0%
momentum is Constant

Explanation

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

$= e[(2i+4j)\times 4k]$

$= e[-8j+16i]$ (By cross-product rule)
So, it has non-zero force, so it velocity direction changes so, path changes, velocity changes, momentum changes but magnitude of velocity is constant, so, speed remains constant.

If a charged particle is projected perpendicular to a uniform magnetic field, then
a) it revolves in circular path
b) its K.E. remains constant
c) its momentum remains constant
d) its path is spiral

0%
only a, b are correct
0%
only a, c are correct
0%
only b, d are correct
0%
only a, d are correct

There will be no force experienced if

0%
two parallel wires carry currents in the same direction
0%
two parallel wires carry currents in the opposite direction
0%
a positive charge is projected between the pole pieces of a bar magnet
0%
a positive charge is projected along the axis of a solenoid carrying current

Explanation

Two parallel wires carrying currents in same direction attract each other.
Two parallel wires carrying currents in opposite direction repel each other.
A positive charge is projected between the pole pieces of bar magnet feels a force =

$q(\vec{V}\times \vec{B})$ which is non zero since angle between V and B is non-zero.
A positive charge projected along the axis of a solenoid carrying current feels no force because in solenoid B is along the axis so

$\vec{V}\times \vec{B}$ is zero.

The coil of the moving coil galvanometer is wound over an aluminium frame

0%
because aluminium is a good conductor
0%
because aluminium is very light.
0%
because aluminium is comparatively cheaper
0%
to provide electro-magnetic damping.

A circular current carrying coil has a radius r. Its magnetic dipole moment is proportional to

0%
r
0%
$r^{2}$
0%
1/r
0%
1/ $r^{2}$

Explanation

Magnetic moment

$= niA$

$= ni \pi r^2$

$(\because A = \pi r^2)$
So,

$M \alpha \ \ r^2$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 1 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 1 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.