NCERT Exemplar Class 12 Physics Solutions Chapter 4 Moving Charges and Magnetism

NCERT Exemplar Class 12 Physics Solutions Chapter 4 MCQ I

Question:1

Two charged particles traverse identical helical paths in an opposite sense in a uniform magnetic field
$B=B_{0}\widehat{k}$
a) they have equal z-components of momenta
b) they must have equal charges
c) they necessarily represent a particle-antiparticle pair
d) the charge to mass ratio satisfy: $\left (\frac{e}{m} \right )_{1}+\left (\frac{e}{m} \right )_{2}=0$

Answer:

The answer is the option (d) As shown in the diagram if the particle is drawn in x-y plane at an angle of θ and moving with a velocity v, so as to make one component along the field $(v \cos\theta )$ and the other perpendicular$(v \sin \theta )$, we need to resolve the velocity in rectangular components. After doing so, the particle gains a constant velocity along the field of $(v \cos\theta )$. The distance travelled by the particle is known as pitch.

The pitch of helix (i.e, linear distance travelled in one rotation ) will be given by $p = T(v\cos \theta )=2p\frac{m}{qB}(v \cos \theta)$
For given pitch p corresponds to change particle, we have
$\frac{q}{m}=\frac{2 \pi v\cos \theta}{qB}$

In this particular case, the path covered by the particles is identical and helical in a completely opposite manner in the presence of a uniform magnetic field, B. This indicates their LHS is the same and has the opposite sign. Therefore, $\left (\frac{e}{m} \right )_{1}+\left (\frac{e}{m} \right )_{2}=0$

Question:2

Biot-Savart law indicates that the moving electrons produce a magnetic field B such that
a) $B\perp v$
b) $B \parallel v$
c) it obeys inverse cube law
d) it is along the line joining the electrons and point of observation

Answer:

The answer is the option (a)
(a) According to the Biot-Savart law , the magnitude of $\overrightarrow{B}$ is $B \propto \left | q \right |, B \propto v$
$B \propto \sin \phi , B \propto \frac{1}{r^{2}}$
$B\propto \frac{\left | q \right | v \sin \phi }{r^{2}}$
$B= \frac{ \mu _{0}}{4 \pi }\frac{\left | q \right | v \sin \phi }{r^{2}}$
Where $\frac{ \mu _{0}}{4 \pi }$ is the proportionality constant 'r' is the magnitude of position vector from charge to the point at which we have to find the magnetic field and $\phi$ is the angle between $\overrightarrow{v}$ and $\overrightarrow{r}$
B is perpendicular to both v and r.

Question:3

A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x >0 is now bent so that it now lies in the y-z plane.
(a) The magnitude of a magnetic moment now diminishes.
(b) The magnetic moment does not change.
(c) The magnitude of B at (0,0, z), z>>R increases.
(d) The magnitude of B at (0,0,z), z>>R is unchanged.

Answer:

The answer is the option (a) According to the right-hand thumb rule, the direction of the magnetic moment of the circular loop is perpendicular to the loop. So, in order to compare their magnetic moments, we need to analyse them in vertical filed.

As shown in the above figure-(a), the circular loop placed in the x-y axis has its magnetic moment across z-axis . And if the loop is half bent with x > 0, it lies in the y-z plane as shown below.

The magnitude of the magnetic moment of each semicircular loop of radius R lies in the x-y plan,e and y-z plane is
$M_{1}= M_{2}= I \frac{\pi R^{2}}{2}$
and the direction of the magnetic moments are along the z-direction and x-direction, respectively. Their resultant is
$M_{net}=\sqrt{M_{1}^{2}+M_{2}^{2}}=\sqrt{2I}\frac{\pi R^{2}}{2}=\frac{M}{\sqrt{2}}$
So M_net < M or M diminishes.

Question:4

An electron is projected with uniform velocity along the axis of a current-carrying long solenoid. Which of the following is true?
a) the electron will be accelerated along the axis
b) the electron path will be circular about the axis
c) the electron will experience a force at 45o to the axis and hence execute a helical path
d) the electron will continue to move with uniform velocity along the axis of the solenoid

Answer:

The answer is the option (d), the electron will continue to move with uniform velocity along the axis of the solenoid.
If the magnetic field of a current-carrying solenoid has a moving electron with uniform velocity across the axis, the electron will face a magnetic force due to magnetic field determined by $F= -evB \sin 180^{\circ} = 0$ (in case of F=0, the velocity is parallel to the magnetic field or anti-parallel or $0=0^{\circ}or180^{\circ}$). It will either result in the electron moving with uniform velocity or it will go undeflected across the solenoid axis.

Question:5

In a cyclotron, a charged particle
(a) Undergoes acceleration all the time
(b) speeds up between the dees because of the magnetic field
(c) speeds up in a dee
(d) slows down within a dee and speeds up between dees

Answer:

The answer is option (a) undergoes acceleration all the time.

Question:6

A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by $30^{\circ}$ about an axis perpendicular to its plane is
(a) $M B$.
(b) $\sqrt{3} \frac{M B}{2}$.
(c) $\frac{M B}{2}$.
(d) zero.

Answer:

The rotation of the loop by $30^{\circ}$ about an axis perpendicular to its plane make no change in the angle made by the axis of the loop with the direction of the magnetic field , therefore, the work done to rotate the loop is zero.

NCERT Exemplar Class 12 Physics Solutions Chapter 4 MCQ II

Question:7

The gyro-magnetic ratio of an electron in an H-atom, according to the Bohr model, is
a) independent of which orbit it is in
b) negative
c) positive
d) increases with the quantum number n

Answer:

The correct answers are the options (a, b) as according to the Bohr model, the gyro-magnetic ration of an electron in an H-atom is negative and independent of which orbit it is in.

Question:8

Consider a wire carrying a steady current, I placed un a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do not work. This implies that
a) motion of charges inside the conductor is unaffected by B since they do not absorb energy
b) some charges inside the wire move to the surface as a result of B
c) if the wire moves under the influence of B, no work is done by the force
d) if the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire

Answer:

The correction options are (b, d).
The force experienced by a straight current-carrying conductor of length (l) at an angle $\theta$ with the direction of the field placed in a uniform magnetic field (B) is Fmax= Bil sin θ. And the direction is determined by the right-hand palm rule.

Question:9

Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C,
(a) $\oint Bdl =\mp\mu _{0}I$
(b) the value of $\oint Bdl =2\mu _{0}I$ is independent of sense of C
(c) there may be a point on C where, B and dl are perpendicular
(d) B vanishes everywhere on C

Answer:

The correct answers are the options (b, c).
In case of the given current distribution, Ampere’s law states another method to calculate the magnetic field.

Question:10

A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v, and a positron enters via the opposite face with velocity -v. At this instant,
(a) the electric forces on both the particles cause identical accelerations.
(b) the magnetic forces on both the particles cause equal accelerations.
(c) both particles gain or lose energy at the same rate.
(d) the motion of the Centre of Mass (CM) is determined by B alone.

Answer:

The correct options are (b, c, d).
$F_{net} =qE + q(v \times B).$
(i) In this case both particles are moving with same acceleration either due to the magnetic forces $(F_m =q(v \times B))$ being zero or the Fm is perpendicular to v (or component of v). This results in the particle moving with a uniform speed in a circular path.
(ii) Both the particle gain or loss energy at the same rate as the electric force is the same (Fe = qE) and in opposite directions due to the sign of charge.
(iii) Th emotion of the centre of mass is determined only by the magnetic field ‘B’ as no change is observed in the centre of mass (CM).

Question:11

A charged particle would continue to move with a constant velocity in a region wherein,
a) $E = 0, B \neq 0$
b) $E \neq 0, B\neq 0$
c) $E \neq 0, B= 0$
d)$E = 0, B = 0$

Answer:

The correct answers are the options (a, b, d).
$F_{net} =qE + q(v \times B)$.
The force created by the electric field on the particle = qE
The force created by the magnetic field on the article $F_{m}= q(v \times B)$
As per the question, the acceleration of the particle is zero due to the constant velocity, and it is not changing its direction of motion. So, if the net force is zero -
(i) if E = 0, and $v\parallel B$, then $F_{net} = 0.$
(ii) if$E \neq 0, B\neq 0$ and E, v and B are mutually perpendicular.
And (iii) when both E and B are absent.

NCERT Exemplar Class 12 Physics Solutions Chapter 4 Very Short Answer

Question:12

Verify that the cyclotron frequency $\omega=e B / m$ has the correct dimensions of $[T]^{-1}$.

Answer:

In the device cyclotron, a circular path is followed by the accelerating particle duet eh magnetic field acting as a centripetal force.
$\frac{mv^{2}}{R}=evB$
On simplifying the terms, we have
$\therefore \frac{eB}{m}=\frac{v}{R}=\omega$
We know that
$B = \frac{F}{eV}=\frac{[MLT^{2}]}{[AT][LT^{-1}]}=[MA^{-1}T^{-2}]$
And then dimensional formula of the angular frequency
$[\omega]= \left [\frac{eB}{m} \right ]=\left [\frac{v}{R} \right ]$
$[\omega ]=\frac{[AT][MA^{-1}T^{-2}]}{[M]}=[T^{-1}]$

Question:13

Show that a force that does no Work must be a velocity-dependent force.

Answer:

To prove that a force with doing no works must be a velocity-dependent force, it is necessary to assume that zero work is done by the force. As given in the following equation:
$dW=\overrightarrow{F}.\overrightarrow{dl}=0$
We can write $\overrightarrow{dl}=\overrightarrow{v}dt, \text{but}dt \neq 0$
$\Rightarrow \overrightarrow{F}.\overrightarrow{v}dt=0\\ \Rightarrow \overrightarrow{F}.\overrightarrow{v}=0$
So, it is proven that the force F is dependent on the velocity, it implies there is a 90° angle between F and v. So if the velocity changes its direction, simultaneously the direction of the force will also change.

Question:14

The magnetic force depends on v which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that-the net acceleration has a different value in different frames of reference?

Answer:

$F = q(v \times B)$
velocity is dependent on the time of reference. But since the magnetic force is also frame-dependent, it changes from inertial frame to frame.
There is a rise in the net acceleration; however, for inertial frames, it is frame independent

Question:14

Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled.

Answer:

To reverse the polarity of the dees at the same time as it takes the ion to complete the one-half revolution, the frequency of the voltage va of the applied voltage is changed. This requirement is called resonance condition, shown as va = vc
The time period of the radio frequency (rf) field was halved due to the violation of the resonance conditions as the frequency of the radio frequency (rf) field were doubled. Thus, two simultaneous things occur, firstly the particle completes its one-half revolution and the radio frequency also completes the cycle.
Therefore, the particle will alternatively accelerate and deaccelerate. So, the radius of the path covered in both the dees remains the same.

Question:16

Two long wires carrying current $I_{1}$, and $I_{2}$ are arranged as shown in figure. The one carrying current $I_{1}$ is along the x-axis. The other carrying current $I_{2}$ is along a line parallel to the y-axis given by x = 0 and z = d. Find the force exerted at $O_{2}$ because of the wire along the x-axis.

Answer:

Firstly, we need to find the direction of the magnetic field caused by one wire on a specific point on the other wire, then determine the magnetic force on the wire carrying the current.
According to Biot-Savart law, the magnetic field is parallel to dl x r whereas I dl direction is along the direction of flow of current, but it can also be found out by using the right-hand thumb rule.

Here the direction of the magnetic field at $O_2$, due to the wire carrying $\mathrm{I}_1$ current is

$
\vec{B} \| i \overrightarrow{d l} \times \vec{r} \text { or } \hat{i} \times \hat{j}=\widehat{k}
$

So the direction of $\mathrm{O}_2$ is along the Y-direction direction of the magnetic force exerted at $\mathrm{O}_2$ because of the wire along the x axis is

$
\vec{F}=\overrightarrow{l l} \times \vec{B}=\hat{j} \times(-\hat{j})=0
$

So the magnetic field due to the wire is along the $y$-axis. The second wire is along the $y$-axis and hence, the force is zero.

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NCERT Exemplar Class 12 Physics Solutions Chapter 4 Short Answer

Question:17

A current-carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B at the origin.

Answer:

The relation of the magnetic field can easily be found out using the Biot-Savart law

Question:18

A charged particle of charge e and mass m is moving in an electric field E, and a magnetic field B. Construct dimensionless quantities and quantities of dimension $[T]^{-1 }$

Answer:

When a charged particle is moving in both electric and magnetic field, the physical quantities can no help to construct dimensional quantity.
The necessary centripetal force is provided by the magnetic Lorentz force when a charged particle is moving perpendicular to the magnetic field.

$\frac{mv^{2}}{R}=evB$
On simplifying the terms, we have
$\therefore \frac{eB}{m}=\frac{v}{R}=\omega$
We know that $B = \frac{F}{eV}=\frac{[MLT^{2}]}{[AT][LT^{-1}]}=[MA^{-1}T^{-2}]$
And
then dimensional formula of the angular frequency $[\omega]= \left [\frac{eB}{m} \right ]=\left [\frac{v}{R} \right ]$
$[\omega ]=\frac{[AT][MA^{-1}T^{-2}]}{[M]}=[T^{-1}]$

Question:19

An electron enters with a velocity $v = v_0i$ into a cubical region (face parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in a plane parallel to the x-y plane. Suggest a configuration to fields E and B that can lead to it.

Answer:

The magnetic force makes a charged particle revolve in a uniform circular motion in the x-y plane, but the electric field increases the speed of the charged particle along the x-direction. This results in increasing the radius of the circular path and so, the moves in a spiral path.
Take the field present in the region to be B = B0, and electrons enter a cubical region with some velocity (faces parallel to coordinate planes). The force due to the Lorentz force on the electron is given by -
$\overrightarrow{F}=-e\left ( v_{0}\widehat{i}\times B_{0}\widehat{k} \right )=ev_{0}B_{0}\widehat{j}$

This revolves the electron in an x-y plane.
With the increase in the radius of the circular path due to ‘e’ acceleration caused by the electric force$F = eE_0j$, the electron transverses on a spiral path.

Question:20

Do magnetic forces obey Newton’s third law. Verify for two current elements $dl_1 = dl i$ located at the origin and $dl_2 = dl j$ located at(0, R, 0). Both carry current I.

Answer:

To solve the problem, we need to determine the magnetic field's direction caused by one wire at the point on other wire, then calculate the magnetic force on the current-carrying wire.

As according to Biot-Savart's law, idl is the current-carrying element along the direction of the flow, and magnetic field $B$ is parallel to $i d l \times r$.

The direction of the magnetic field is given by $B \| i d l \times \operatorname{or} \hat{i} \times j$ (because point $(0, \mathrm{R}, 0)$ lies ony-axis), but $i \times j=k$ which is present at $i d l_2$ and located at located at $(0, \mathrm{R}, 0)$ due to the wire dlx.

This makes the direction the magnetic field along z-direction at $d l_2$.
And the magnetic field present at dl 2 due to the magnetic field of first wire has its direction in the x -axis.
$F-i(l \times B)$, i.e., $F \|(i \times k)$ or along - j direction.
Therefore, it makes the forces present due to $\mathrm{dl}\left[\right.$ on $d l_2$ is non-zero.
Now, to find he direction of the magnetic field at $d x$ due to the $d_2$ wire located at $(0,0,0)$ is calculated by the relation $B \| i d l \times \operatorname{or} j \times-j$ (because origin lies on y -direction with respect to the point $(0, \mathrm{R}, 0)$, but $j \times-j=0$.

This implies that there is no magnetic field present at dx .
Hence, force due to $d l_2$ on $d l_1$, is zero.
So,Newton's third law is not followed by the magnetic forces. But in case of current carrying elements present parallel to each other they will follow Newton's third law.

Question:21

A multirange voltmeter can be constructed by using a galvanometer circuit as shown in figure. We want to construct a voltmeter that can measure 2 V, 20 V and 200 V using a galvanometer of resistance 10$\Omega$ and that produces maximum deflection for a current of 1 mA. Find $R_1, R_2 and R_3$ that have to be used,

Answer:

The device named galvanometer can also be used to measure the voltage across a given circuit section just like a voltmeter. To do so, a very high resistance wire is connected to the galvanometer in series. The relation is determined by Ig (G + R) – V in which Ig is galvanometer’s range, G is its resistance and R represents the resistance of the wire connected to the galvanometer in series.

Applying the expression in a different situation
For $i_{G}(G+R_{1})=2 $ for 2V range
For $i_{G}(G+R_{1}+R_{2})=20 $ for 20V range
And for $i_{G}(G+R_{1}+R_{2}+R_{3})=200$ for 200V range
By solving we get
$R_{1}=1990 \Omega , R_{2}=18k\Omega$ and $R_{3}=180 k \Omega$

Question:22

A long straight wire carrying a current of 25 A rests on a table as shown in figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

Answer:

The long straight wire carrying 25 A current resting on the table applies a force on PQ. And if two straight wires are placed parallel to the current in opposite directions, the forces are repulsive. But if there is an equilibrium in wire PQ, then the repulsive force on PQ balances its weight

The magnetic field produced by a long straight wire carrying a current of 25A rests on the table on small wire.
$B= \frac{\mu _{0}I}{2 \pi h}$
The magnetic force on the small conductors
$F = BlI \sin \theta =BIl$
Force applied on PQ balances the weight of the small current carrying capacity
$F=mg= \frac{\mu _{0}I^{2}l}{2 \pi h}$
$h= \frac{\mu _{0}I^{2}l}{2 \pi mg}=\frac{4\pi \times 10^{-7}\times (25)^{2}\times l}{2\pi \times 2.5 \times 10^{-3}\times 9.8 }=5l\times 10^{-4}m = 0.51m$

NCERT Exemplar Class 12 Physics Solutions Chapter 4 Long Answer

Question:23

100 turn rectangular coil ABCD (in X-Y plane) is hung from one arm of a balance (shown in figure). A mass of 500 g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil, and a constant magnetic field of 0.2 T acting inward (in x-z plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass m must be added to regain the balance?

Answer:

The concept of the magnetic force on a conductor placed in the region of an external uniform magnetic field is used. The external magnetic field causing the magnetic force on CD must balance its weight.
And the net torque present on the spring balance must be zero to attain equilibrium.
At t=0 ,the external magnetic field is off. Let us consider the sepration of each hung from the mid-point be l.
$Mgl = W_{coil}l$
$0.5gl = W_{coil}l$
$W_{coil}=0.5 \times 9.8$
By taking the moment of the about the mid-point , we get the weigh of the coil. And let 'm' be the mass which is added to regain the balance . When the magnetic field is switched on.
$Mgl +mgl=W_{coil}l+(IlB\sin 90^{\circ})l$
$Mgl =(IlB)l$
$m=\frac{BIL}{g}=\frac{0.2 \times 4.9 \times 1 \times 10^{-2}}{9.8}=10^{-3}kg =1g$
Therefore 1g of additional mass must be added to regain the balance.

Question:24

A rectangular conducting loop consists of two wires on two opposite sides of length l joined together by rods of length d. The wires each are of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance R and the rods are of low resistance, which in turn are connected to a constant voltage source $V_{0}$. The loop is placed in a uniform magnetic field B at$45^{\circ}$ to its plane. Find T, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.

Answer:

To find the net torque it is necessary to calculate the magnetic forces and torques after analysis of the direction of current in both wires.

According to the problem, the thicker wire has the resistance R, then the other wire has resistance 2Ras the wires has resistance 2R as the wires of the same material so their resistivity remains the same.
Now, the force and hence, torque on the first wire is given by
$F_{1}=i_{1}lB\sin 90^{\circ}=\frac{V_{0}}{2R}IB$
$\tau _{1}=\frac{d}{2\sqrt{2}} F_{1} =\frac{V_{0}IdB}{2\sqrt{2}R}$
Similarly, the force hence torque on the other wire is given by

$F_{2}=i_{2}lB\sin 90^{\circ}=\frac{V_{0}}{2R}IB$
$\tau _{2}=\frac{d}{2\sqrt{2}} F_{2} =\frac{V_{0}IdB}{4\sqrt{2}R}$
So, net torque $\tau =\tau_{1}-\tau_{2}$
$\tau =\frac{l}{4\sqrt{2}}\frac{V_{0}AB}{R}$
Where A is the area of the rectangular coil.

Question:25

An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R) respectively, in a uniform magnetic field B = B₀i, each with an equal momentum of magnitude p = eBR. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

Answer:

Due to the presence of the magnetic field B along the x-axis, the momenta of two particles in a circular orbit is in the y-z plane. Assume the momenta of the electron (e– ) and positron (e+) to be p₁ and p₂, respectively. Both moves in the opposite sense in a circle of radius R. Let p, make an angle θ with the y-axis p₂ must make the same angle withy axis.
But the respective circles must have their centres perpendicular to the momenta at a distance R. Let Ce be the centre of the electron and Cp of the positron.
The coordinates of Ce are $Ce = (0, -R \sin \theta, R \cos \theta)$
The coordinates of Cp are $Cp = [0, -R \sin \theta, (1.5R - R \cos \theta)]$

The circular orbits of electron and positron shall not overlap if the distance between the two centres are greater than 2R.
Let d be the distance between Cp and Ce. Then
$d^{2}=[R\sin \theta -(-R \sin \theta )]^{2}\left [ R\cos \theta \left ( \frac{3}{2}R-R\cos \theta \right ) \right ]^{2}$
$=(2R\sin \theta)^{2}+\left ( 2R \cos \theta -\frac{}3{}2 R \right )^{2}$
$=4R^{2}\sin ^{2}\theta +4R^{2}\cos^{2}\theta -6R^{2}\cos \theta +\frac{9}{4}R^{2}$
$=4R^{2} -6R^{2}\cos \theta +\frac{9}{4}R^{2}$
As d has to greater than 2R d² > 4R^{2
$=4R^{2} -6R^{2}\cos \theta +\frac{9}{4}R^{2}>4R^{2}$
$\frac{9}{4}>6 \cos \theta or \cos \theta <\frac{3}{8}$}

Question:26

A uniform conducting wire of length 12a and resistance R is wound up as a current-carrying coil in the shape of (i) an equilateral triangle of side a, (ii) a square of sides a, and (iii) a regular hexagon of sides a. The coil is connected to a voltage source V0. Find the magnetic moment of the coils in each case.

Answer:

(i) For equilateral triangle of side a,

As the total wire length = 12a, the no. of the loops
$n = \frac{12a}{3a}=4$
The magnetic moment of the coils m = nIA
As the area of the triangle is
$A=\frac{\sqrt{3}}{4}a^{2}$
$=4I\left (\frac{\sqrt{3}}{4}a^{2} \right )$
$\therefore m=Ia^{2}\sqrt{3}$
Iii) For a square of sides a,

$A=a^{2}$
No. of loops n =
$\frac{12a}{4a}=3$
Magnetic moment of the coils m =nIA= 3I(a²) = 3Ia²
(iii) For regular hexagon of sides a,

No. of loops n = $\frac{12a}{6a}=2$
Aera,
$A=\frac{6 \sqrt{3}}{4}a^{2}$
Magnetic moment of the coils m = nIA
$\Rightarrow m=2I=\left (\frac{6 \sqrt{3}}{4}a^{2} \right )I$
$\Rightarrow m=3\sqrt{3}a^{2}I$, m is in a geometric series.

Question:27

Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the line integral
$\tau (L)=\left | \int_{-L}^{L} B.dl\right |$ taken along z-axis

(a) Show that $\tau (L)$ monotonically increases with L.
b) Use an appropriate Amperian loop to show that $\tau (\infty )=\mu_{0}I$ , where I is the current in the wire.
(c) Verify directly the above result.
(d) Suppose we replace the circular coil by a square coil of sides R carrying the same current I. What can you say about $\tau (L )$and $\tau (\infty )$

Answer:

(a) In a circular current-carrying loop in x-y plane, the magnetic field acts along z-axis as shown in below.

$\tau (L)=\left | \int_{-L}^{L} B.dl\right |=\int_{-L}^{+L}Bdl \cos 0^{\circ}=\int_{-L}^{+L}Bdl= 2BL$
$\therefore \tau (L)$ is monotonically increasing function of L
(b) Now consider an Amperean loop around the circular coil of such a large radius that $L\rightarrow \infty$. Since this loop encloses a current I, now using the Amperan's law
$\tau (\infty )=\oint_{-\infty }^{+\infty }\overrightarrow{B}.\overrightarrow{dl}=\mu _{0}I$
(c) The magnetic field at the axis (z-axis) of the circular coil is given by
$B){z}=\frac{\mu _{0}IR^{2}}{2(z^{2}+R^{2})^{\frac{3}{2}}}$
Now intergarting
$\int_{-\infty }^{+\infty }B_{z}=\int_{-\infty }^{+\infty }\frac{\mu _{0}IR^{2}}{2(z^{2}+R^{2})^{\frac{3}{2}}}dz$
Let z = R tan $\theta \ so \ that \ dz = r\sec ^{2}\theta d\theta$
and
$(z^{2}+R^{2})^{\frac{3}{2}}=(R^{2}\tan ^{2}\theta +R^{2})^{\frac{3}{2}}\\ = R^{3}\sec^{3}\theta(as 1 +\tan ^{2}\theta \sec^{2}\theta)$
Thus,
$\int_{-\infty }^{+\infty }B_{z}=\frac{\mu _{0}I}{2}\int_{-\frac{\pi}{2} }^{+\frac{\pi} {2}}\frac{R^{2}(R\sec ^{2}\theta d \theta)}{R^{3}\sec ^{3}\theta }$
$=\frac{\mu _{0}I}{2}\int_{-\frac{\pi}{2} }^{+\frac{\pi} {2}}\cos \theta d \theta =\mu _{0}I$
(d) As we know $(B_{z})_{square}<(B_{z})_{circular coil}$
For the same current and side of the square equal to radius of the coil
$\tau(\infty )_{square}<\tau(\infty )_{circular coil}$
By using the same argument as we done in the case (b), it can be shown that
$\tau(\infty )_{square}=\tau(\infty )_{circular coil}$

Question:28

A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure 10 mA, 100 mA and 1 mA using a galvanometer of resistance $10 \Omega$and that produces maximum deflection for current of 1 mA. Find $S_{1}. S_{2}$ and $S_{3}$ that have to be used.

Answer:

A galvanometer is sometimes also used as an ammeter by simply attaching a very low resistance wire also called as shunt S in parallel to the galvanometer. IgG = (I – Ig) S gives the relationship, in which range of galvanometer is Ig and R is the resistance of the galvanometer.

For measuring $I _{1}=10mA: I_{G}.G=(I_{1}-I_{G})(S_{1}+S_{2}+S_{3})$
For measuring $I _{2}=100mA: I_{G}.(G+S_{1})=(I_{2}-I_{G})(S_{2}-S_{3})$
For measuring $I _{3}=1A: I_{G}.(G+S_{1}+S_{2})=(I_{3}-I_{G})(S_{3})$
gives $S_{1}= 1\Omega , S_{2}= 0.1 \Omega , S_{3}= 0.01 \Omega$

Main Subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 4 Moving Charges and Magnetism

Below are the main subtopics covered in NCERT Exemplar Class 12 Physics Chapter 4 – Moving Charges and Magnetism, which build a strong foundation for understanding the magnetic effects of electric current.

• Introduction
• Magnetic Force
• Sources And Fields
• Magnetic Field, Lorentz Force
• Magnetic Force On A Current-Carrying Conductor
• Motion In A Magnetic Field
• Motion In Combined Electric And Magnetic Fields
• Velocity Selector
• Cyclotron
• Magnetic Field Due To A Current Element, Biot-Savart Law
• Magnetic Field On The Axis Of A Circular Current Loop
• Ampere's Circuital Law
• The Solenoid And The Toroid
• The Solenoid
• The Toroid
• Force Between Two Parallel Currents, The Ampere
• Torque On Current Loop, Magnetic Dipole
• Torque On A Rectangular Current Loop In A Uniform Magnetic Field
• Circular Current Loop As A Magnetic Dipole
• The Magnetic Dipole Moment Of A Revolving Electron
• The Moving Coil Galvanometer

NCERT Exemplar Class 12 Physics Chapter Wise Links

NCERT Exemplar Class 12 Solutions

Also, check the NCERT solutions of questions given in the book

Also, read NCERT Solutions Class 12 subject-wise

• NCERT Solution for Class 12 Physics
• NCERT Solution for Class 12 Chemistry
• NCERT Solution for Class 12 Maths
• NCERT Solution for Class 12 Biology

Must read NCERT Notes subject wise

• NCERT Notes for Class 12 Physics
• NCERT Notes for Class 12 Chemistry
• NCERT Notes for Class 12 Maths
• NCERT Notes for Class 12 Biology

Also, check the NCERT Books and NCERT Syllabus here:

• NCERT Books Class 12 Physics
• NCERT Syllabus Class 12 Physics
• NCERT Books Class 12
• NCERT Syllabus Class 12