Learn › NEET · Physics PYQ › Moving Charges and Magnetism
Moving Charges and Magnetism — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Moving Charges and Magnetism MCQs with step-by-step solutions covering Biot Savart's Law and Ampere's Circuital Law, Magnetic Force on Charged Particle in Magnetic Field, Force and Torque on Current Carrying Conductor, Moving Coil Galvanometer. Practise online on Prepizo — no login needed.
▶ Practise Moving Charges and Magnetism online (free)
Subtopics
Sample questions with solutions
Q1 — Biot Savart's Law and Ampere's Circuital Law · easy · numerical
A long solenoid of $50\,\text{cm}$ length having $100$ turns carries a current of $2.5\,\text{A}$. The magnetic field at the centre of solenoid is (Take, $\mu_0=4\pi\times10^{-7}\,\text{T m A}^{-1}$)
A. $3.14\times10^{-4}\,\text{T}$
B. $6.28\times10^{-5}\,\text{T}$
C. $3.14\times10^{-5}\,\text{T}$
D. $6.28\times10^{-4}\,\text{T}$ ✓ Correct
Solution: $B=\mu_0 n I=\mu_0\left(\dfrac{N}{l}\right)I=4\pi\times10^{-7}\times\dfrac{100}{0.5}\times2.5=6.28\times10^{-4}\,\text{T}$.
Q2 — Biot Savart's Law and Ampere's Circuital Law · easy · theory
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is $B$. It is then bent into a circular coil of $n$ turns. The magnetic field at the centre of this coil of $n$ turns will be
A. $nB$
B. $n^2B$ ✓ Correct
C. $2nB$
D. $2n^2B$
Solution: Since the wire length is fixed, the radius becomes $R/n$ for $n$ turns. $B_{centre}=\dfrac{n\mu_0 i}{2r}$; substituting $r=R/n$ gives $B_{new}=n^2 B$.
Q3 — Biot Savart's Law and Ampere's Circuital Law · easy · numerical
Magnetic field due to $0.1\,\text{A}$ current flowing through a circular coil of radius $0.1\,\text{m}$ and $1000$ turns at the centre of the coil is
A. $0.2\,\text{T}$
B. $2\pi\times10^{-4}\,\text{T}$
C. $6.28\times10^{-4}\,\text{T}$ ✓ Correct
D. $9.8\times10^{-4}\,\text{T}$
Solution: $B=\dfrac{\mu_0 N i}{2r}=\dfrac{4\pi\times10^{-7}\times1000\times0.1}{2\times0.1}=2\pi\times10^{-4}=6.28\times10^{-4}\,\text{T}$.
Q4 — Biot Savart's Law and Ampere's Circuital Law · easy · theory
If a long hollow copper pipe carries a current, then magnetic field is produced
A. inside the pipe only
B. outside the pipe only ✓ Correct
C. both inside and outside the pipe
D. no where
Solution: By Ampere's circuital law $\oint \vec{B}\cdot d\vec{l}=\mu_0 i_{\text{enclosed}}$; inside the hollow pipe $i_{\text{enclosed}}=0$ so $B=0$, while outside the pipe the current behaves as if concentrated on the axis, giving $B=\dfrac{\mu_0 i}{2\pi r}\neq 0$.
Q5 — Biot Savart's Law and Ampere's Circuital Law · easy · theory
A straight wire of diameter $0.5\,\text{mm}$ carrying a current of $1\,\text{A}$ is replaced by another wire of $1\,\text{mm}$ diameter carrying the same current. The strength of magnetic field far away is
A. twice the earlier value
B. same as the earlier value ✓ Correct
C. one-half of the earlier value
D. one-quarter of the earlier value
Solution: The field far from a straight wire, $B=\dfrac{\mu_0}{4\pi}\dfrac{2i}{r}$, depends only on the current $i$ and the distance $r$ from the wire, not on the wire's diameter, so the field is unchanged.
Q6 — Biot Savart's Law and Ampere's Circuital Law · easy · numerical
The magnetic field at a distance $r$ from a long wire carrying current $i$ is $0.4\,\text{T}$. The magnetic field at a distance $2r$ is
A. $0.2\,\text{T}$ ✓ Correct
B. $0.8\,\text{T}$
C. $0.1\,\text{T}$
D. $1.6\,\text{T}$
Solution: Since $B\propto \dfrac{1}{r}$, doubling the distance halves the field: $B'=\dfrac{0.4}{2}=0.2\,\text{T}$.
Q7 — Biot Savart's Law and Ampere's Circuital Law · easy · numerical
The magnetic induction at a point $P$ which is at a distance of $4\,\text{cm}$ from a long current carrying wire is $10^{-3}\,\text{T}$. The field of induction at a distance $12\,\text{cm}$ from the current will be
A. $3.33\times10^{-4}\,\text{T}$ ✓ Correct
B. $1.11\times10^{-4}\,\text{T}$
C. $3\times10^{-3}\,\text{T}$
D. $9\times10^{-3}\,\text{T}$
Solution: Since $B\propto \dfrac{1}{r}$, tripling the distance from $4\,\text{cm}$ to $12\,\text{cm}$ reduces the field to one-third: $B'=\dfrac{10^{-3}}{3}=3.33\times10^{-4}\,\text{T}$.
Q8 — Biot Savart's Law and Ampere's Circuital Law · easy · theory
Tesla is the unit of
A. magnetic flux
B. magnetic field
C. magnetic induction ✓ Correct
D. magnetic moment
Solution: The SI unit of magnetic induction (magnetic flux density) is the tesla (T): $1\,\text{T}=1\,\text{N}\,\text{A}^{-1}\,\text{m}^{-1}$, the field that exerts a force of $1\,\text{N}$ on a charge of $1\,\text{C}$ moving at $1\,\text{m/s}$ perpendicular to it.
Q9 — Magnetic Force on Charged Particle in Magnetic Field · easy · theory
A particle of mass $m$, charge $q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $B$. After $3\ \text{s}$, the kinetic energy of the particle will be
A. $3T$
B. $2T$
C. $T$ ✓ Correct
D. $4T$
Solution: The magnetic force is always perpendicular to velocity, so it does no work and never changes the kinetic energy of the particle. Hence the kinetic energy remains $T$.
Q10 — Magnetic Force on Charged Particle in Magnetic Field · easy · theory
A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched OFF and the same magnetic field is maintained, the electrons move
A. in an elliptical orbit
B. in a circular orbit ✓ Correct
C. along a parabolic path
D. along a straight line
Solution: With the electric field off, only the magnetic force (perpendicular to velocity) acts on the electrons, causing them to move in a circular orbit.
Q11 — Magnetic Force on Charged Particle in Magnetic Field · easy · theory
Under the influence of a uniform magnetic field, a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion
A. depends on $v$ and not on $R$
B. depends on both $R$ and $v$
C. is independent of both $R$ and $v$ ✓ Correct
D. depends on $R$ and not on $v$
Solution: Since $T = \dfrac{2\pi m}{qB}$, the time period depends only on the mass, charge and magnetic field, and is independent of both $R$ and $v$.
Q12 — Magnetic Force on Charged Particle in Magnetic Field · easy · theory
When a charged particle moving with velocity $\vec{v}$ is subjected to a magnetic field of induction $\vec{B}$, the force on it is non-zero. This implies that
A. angle between $\vec{v}$ and $\vec{B}$ is necessarily $90^\circ$
B. angle between $\vec{v}$ and $\vec{B}$ can have any value other than $90^\circ$
C. angle between $\vec{v}$ and $\vec{B}$ can have any value other than zero and $180^\circ$ ✓ Correct
D. angle between $\vec{v}$ and $\vec{B}$ is either zero or $180^\circ$
Solution: Since $F = qvB\sin\theta$, a non-zero force requires $\sin\theta \neq 0$, i.e., $\theta$ can be any value other than $0^\circ$ and $180^\circ$.
Q13 — Magnetic Force on Charged Particle in Magnetic Field · easy · theory
A charged particle moves through a magnetic field in a direction perpendicular to it. Then, the
A. acceleration remains unchanged
B. velocity remains unchanged
C. speed of the particle remains unchanged ✓ Correct
D. direction of the particle remains unchanged
Solution: The magnetic force continuously changes the direction of the particle but never does work on it, so the speed (magnitude of velocity) remains unchanged.
Q14 — Magnetic Force on Charged Particle in Magnetic Field · easy · theory
A charge $q$ moves in a region where electric field $\vec{E}$ and magnetic field $\vec{B}$ both exist, then the force on it is
A. $q(\vec{v}\times\vec{B})$
B. $q\vec{E} + q(\vec{v}\times\vec{B})$ ✓ Correct
C. $q\vec{B} + q(\vec{B}\times\vec{v})$
D. $q\vec{B} + q(\vec{E}\times\vec{v})$
Solution: The net Lorentz force on the charge is the sum of the electric force and the magnetic force: $\vec{F} = q\vec{E} + q(\vec{v}\times\vec{B})$.
Q15 — Magnetic Force on Charged Particle in Magnetic Field · easy · theory
A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will
A. continue to move due East
B. move in a circular orbit with its speed unchanged ✓ Correct
C. move in a circular orbit with its speed increased
D. get deflected vertically upwards
Solution: The magnetic field is perpendicular to the velocity, so the magnetic force $F=q\vec{v}\times\vec{B}$ always acts perpendicular to $\vec{v}$, doing no work. The particle moves in a circular orbit with unchanged speed.
Q16 — Magnetic Force on Charged Particle in Magnetic Field · easy · numerical
A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength $20\ \text{Vm}^{-1}$ and $0.5\ \text{T}$ respectively, at right angles to the direction of motion of the electrons. Then, the velocity of electrons must be
A. $8\ \text{m/s}$
B. $20\ \text{m/s}$
C. $40\ \text{m/s}$ ✓ Correct
D. $\dfrac{1}{40}\ \text{m/s}$
Solution: For undeviated motion (velocity selector condition), $qE=qvB \Rightarrow v=\dfrac{E}{B}=\dfrac{20}{0.5}=40\ \text{m/s}$.
Q17 — Magnetic Force on Charged Particle in Magnetic Field · easy · theory
A charge moving with velocity $v$ in the $x$-direction is subjected to a field of magnetic induction in the negative $x$-direction. As a result, the charge will
A. remain unaffected ✓ Correct
B. start moving in a circular $y$-$z$ plane
C. retard along $x$-axis
D. move along a helical path around the $x$-axis
Solution: Force $F=qvB\sin\theta$; here $\vec{v}$ and $\vec{B}$ are anti-parallel (angle $180^\circ$), so $\sin 180^\circ=0$ and $F=0$. The charge remains unaffected.
Q18 — Magnetic Force on Charged Particle in Magnetic Field · easy · numerical
A uniform magnetic field acts at right angles to the direction of motion of electrons. As a result, the electron moves in a circular path of radius 2 cm. If the speed of the electrons is doubled, the radius of the circular path will be
A. $2.0\ \text{cm}$
B. $0.5\ \text{cm}$
C. $4.0\ \text{cm}$ ✓ Correct
D. $1.0\ \text{cm}$
Solution: Since $r=\dfrac{mv}{qB}$, $r\propto v$. Doubling the speed doubles the radius: $r'=2\times 2=4\ \text{cm}$.
Q19 — Force and Torque on Current Carrying Conductor · easy · theory
Current is flowing in a coil of area $A$ and number of turns $N$, then magnetic moment of the coil, $M$ is equal to
A. $NiA$ ✓ Correct
B. $\dfrac{Ni}{A}$
C. $\dfrac{Ni}{\sqrt{A}}$
D. $N^2Ai$
Solution: The magnetic dipole moment (magnetic moment) of a current-carrying coil with $N$ turns, current $i$, and area $A$ is $M=NiA$.
Q20 — Force and Torque on Current Carrying Conductor · easy · numerical
Two long parallel wires are at a distance of $1\,\text{m}$. Both of them carry $1\,\text{A}$ of current. The force of attraction per unit length between the two wires is
A. $2\times10^{-7}\,\text{N/m}$ ✓ Correct
B. $2\times10^{-8}\,\text{N/m}$
C. $5\times10^{-8}\,\text{N/m}$
D. $10^{-7}\,\text{N/m}$
Solution: $\dfrac{F}{l}=\dfrac{\mu_0}{2\pi}\dfrac{i_1i_2}{r}=\dfrac{4\pi\times10^{-7}}{2\pi}\times\dfrac{1\times1}{1}=2\times10^{-7}\,\text{N/m}$.
Q21 — Force and Torque on Current Carrying Conductor · easy · numerical
A straight wire of length $0.5\,\text{m}$ and carrying a current of $1.2\,\text{A}$ is placed in uniform magnetic field of induction $2\,\text{T}$. The magnetic field is perpendicular to the length of the wire. The force on the wire is
A. $2.4\,\text{N}$
B. $1.2\,\text{N}$ ✓ Correct
C. $3.0\,\text{N}$
D. $2.0\,\text{N}$
Solution: $F=BiL\sin\theta$, with $\theta=90^\circ$ (field perpendicular to wire): $F=2\times1.2\times0.5=1.2\,\text{N}$.
Q22 — Force and Torque on Current Carrying Conductor · easy · theory
A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes
A. inclined at $45^\circ$ to the magnetic field
B. inclined at any arbitrary angle to the magnetic field
C. parallel to the magnetic field ✓ Correct
D. perpendicular to the magnetic field
Solution: The coil orients itself so that its magnetic moment $\vec M$ becomes parallel to $\vec B$, which means the plane of the coil becomes parallel to the field, giving zero net torque (stable equilibrium).
Q23 — Biot Savart's Law and Ampere's Circuital Law · hard · numerical
When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards West. When it is projected towards North with a speed $v_0$, it moves with an initial acceleration $3a_0$ towards West. The electric and magnetic fields in the room are
A. $\dfrac{ma_0}{e}$ West, $\dfrac{2ma_0}{ev_0}$ up
B. $\dfrac{ma_0}{e}$ West, $\dfrac{2ma_0}{ev_0}$ down ✓ Correct
C. $\dfrac{ma_0}{e}$ East, $\dfrac{3ma_0}{ev_0}$ up
D. $\dfrac{ma_0}{e}$ East, $\dfrac{3ma_0}{ev_0}$ down
Solution: From rest, only the electric force acts: $a_0=\dfrac{eE}{m}\Rightarrow E=\dfrac{ma_0}{e}$ West. Moving North with $v_0$, the magnetic force adds: $ev_0B+eE=3ma_0\Rightarrow B=\dfrac{2ma_0}{ev_0}$, directed vertically downward.
Q24 — Biot Savart's Law and Ampere's Circuital Law · hard · numerical
Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potential difference applied across them so that the magnetic field at their centres is the same?
A. $3$
B. $4$ ✓ Correct
C. $6$
D. $2$
Solution: $B=\dfrac{\mu_0 i}{2r}$; equal $B$ needs $\dfrac{i_1}{r_1}=\dfrac{i_2}{r_2}$, and with $r_1=2r_2$, $i_1=2i_2$. Since resistance $\propto$ length $\propto r$, $\dfrac{V_1}{V_2}=\dfrac{i_1 r_1}{i_2 r_2}=\dfrac{2i_2(2r_2)}{i_2 r_2}=4$.
Q25 — Magnetic Force on Charged Particle in Magnetic Field · hard · theory
In the product $\vec{F} = q(\vec{v} \times \vec{B}) = q\vec{v} \times (B\hat{i} + B\hat{j} + B_0\hat{k})$. For $q = 1$ and $\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$, what will be the complete expression for $\vec{B}$?
A. $-8\hat{i}-8\hat{j}-6\hat{k}$
B. $-6\hat{i}-6\hat{j}-8\hat{k}$ ✓ Correct
C. $8\hat{i}+8\hat{j}-6\hat{k}$
D. $6\hat{i}+6\hat{j}-8\hat{k}$
Solution: Taking $\vec{v} \times \vec{B}$ with $\vec{B} = B\hat{i}+B\hat{j}+B_0\hat{k}$ and equating components to $\vec{F}=4\hat{i}-20\hat{j}+12\hat{k}$ gives $2B-4B=12 \Rightarrow B=-6$, and $6B-2B_0=-20 \Rightarrow B_0=-8$. So $\vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}$.
Q26 — Magnetic Force on Charged Particle in Magnetic Field · hard · theory
An alternating electric field of frequency $\nu$ is applied across the dees (radius $= R$) of a cyclotron that is being used to accelerate protons (mass $= m$). The operating magnetic field $B$ used in the cyclotron and the kinetic energy $K$ of the proton beam produced by it are given by
A. $B = \dfrac{m\nu}{e}$ and $K = 2m\pi^2\nu^2R^2$
B. $B = \dfrac{2\pi m\nu}{e}$ and $K = m^2\pi\nu R^2$
C. $B = \dfrac{2\pi m\nu}{e}$ and $K = 2m\pi^2\nu^2R^2$ ✓ Correct
D. $B = \dfrac{m\nu}{e}$ and $K = m^2\pi\nu R^2$
Solution: Cyclotron frequency $\nu = \dfrac{eB}{2\pi m} \Rightarrow B = \dfrac{2\pi m\nu}{e}$. Maximum speed $v = 2\pi R \nu$, so $K = \dfrac{1}{2}mv^2 = 2m\pi^2\nu^2R^2$.
Q27 — Magnetic Force on Charged Particle in Magnetic Field · hard · numerical
The magnetic force acting on a charged particle of charge $-2\ \mu\text{C}$ in a magnetic field of $2\ \text{T}$ acting in the $y$-direction, when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^{6}\ \text{ms}^{-1}$, is
A. $8\ \text{N}$ in $-z$-direction ✓ Correct
B. $4\ \text{N}$ in $-z$-direction
C. $8\ \text{N}$ in $+y$-direction
D. $8\ \text{N}$ in $+z$-direction
Solution: $\vec{F} = q(\vec{v}\times\vec{B}) = (-2\times10^{-6})\{(2\hat{i}+3\hat{j})\times10^6 \times 2\hat{j}\} = (-2\times10^{-6})[4\times10^6\hat{k}] = -8\ \text{N}\hat{k}$, i.e., $8\ \text{N}$ along the negative $z$-axis.
Q28 — Force and Torque on Current Carrying Conductor · hard · numerical
A metallic rod of mass per unit length $0.5\ \text{kg m}^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^\circ$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25\ \text{T}$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
A. $14.76\ \text{A}$
B. $5.98\ \text{A}$
C. $7.14\ \text{A}$
D. $11.32\ \text{A}$ ✓ Correct
Solution: For equilibrium along the incline, $mg\sin\theta = F\cos\theta = ILB\cos\theta$, giving $I=\dfrac{m}{L}\dfrac{g\tan\theta}{B}=\dfrac{0.5\times 9.8\times\tan 30^\circ}{0.25}=\dfrac{0.5\times 9.8}{0.25\sqrt{3}}\approx 11.32\ \text{A}$.
Q29 — Force and Torque on Current Carrying Conductor · hard · theory
A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $\vec{F}$ is in the plane of the coil. If due to a current $i$ in the triangle a torque $\tau$ acts on it, the side $l$ of the triangle is
A. $\dfrac{2}{\sqrt3}\left(\dfrac{\tau}{Bi}\right)^{1/2}$
B. $\dfrac{2}{\sqrt3}\left(\dfrac{\tau}{Bi}\right)$
C. $2\left(\dfrac{\tau}{\sqrt3\,Bi}\right)^{1/2}$ ✓ Correct
D. $\dfrac{1}{\sqrt3}\dfrac{\tau}{Bi}$
Solution: Torque on the coil is $\tau = iAB\sin90^\circ = iAB$, and the area of an equilateral triangle of side $l$ is $A=\dfrac{\sqrt3}{4}l^2$. So $\tau=\dfrac{\sqrt3}{4}l^2Bi \Rightarrow l=2\left(\dfrac{\tau}{\sqrt3\,Bi}\right)^{1/2}$.
Q30 — Moving Coil Galvanometer · hard · numerical
In an ammeter $0.2\%$ of main current passes through the galvanometer. If resistance of galvanometer is $G$, the resistance of the ammeter will be
A. $\dfrac{1}{499}G$
B. $\dfrac{499}{500}G$
C. $\dfrac{1}{500}G$ ✓ Correct
D. $\dfrac{500}{499}G$
Solution: With $0.002I$ through the galvanometer $G$ and $0.998I$ through the shunt $r_s$: $0.002IG=0.998Ir_s \Rightarrow r_s\approx\dfrac{G}{499}$. The ammeter's equivalent resistance is $\dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{r_s}=\dfrac{1}{G}+\dfrac{499}{G}=\dfrac{500}{G}\Rightarrow R=\dfrac{G}{500}$.