Magnetism Previous Year Questions With Solutions

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Magnetism Previous Year Questions With Solutions for JEE Mains

Magnetism Previous Year Questions With Solutions

1.0Introduction

Moving charges produce magnetic fields, which is the basic principle behind the magnetic effect of electric current.Oersted experiment discovered that a current-carrying wire generates a magnetic field around it, linking electricity and magnetism. According to the Right-Hand Thumb Rule, the direction of the magnetic field can be determined by pointing your thumb in the direction of current and curling your fingers. When current flows through a coil (solenoid), it creates a magnetic field like that of a bar magnet—this is called electromagnetism. Magnetic field lines form closed loops from north to south outside the magnet and never intersect. Materials like iron, cobalt, and nickel show magnetism due to aligned atomic magnetic moments. This phenomenon is widely used in motors, generators, electromagnets, and various electronic devices.

JEE Main Previous Year Solved Questions on Magnetism
JEE Adv Previous Year Solved Questions on Magnetism

2.0Key Concepts to Remember

Concept of Field and Oersted Experiment

A region around any physical quantity where another similar physical quantity experiences force or torque is called a field.

Oersted Experiment

When the direction of current in the conductor is reversed then deflection of the magnetic needle is also reversed.
Increasing current or moving the needle closer increases its deflection.

Biot Savart Law

The magnetic field at a point is directly proportional to the current, element length, and sine of the angle, and inversely proportional to the square of the distance.

Ampere’s Circuital Law

Line integral of magnetic field along any closed loop is equal to o times the net current crossing the surface bounded by the loop.

Magnetic field due to Solenoid

It is a wire wound into a helix with insulated turns. The magnetic field inside is strong and aligned along the axis, while outside it is nearly zero. The field direction inside is determined using the right-hand thumb rule.

Magnetic field due to Toroid

A toroid can be considered as a ring-shaped closed solenoid also called endless solenoid.Toroid is a solenoid bent in ring shape.

Bar Magnet: It is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, having permanent magnetic properties. Two poles are present in a Bar magnet, North Pole and South Pole.

Moving Coil Galvanometer

The galvanometer has a coil with many turns, free to rotate in a uniform radial magnetic field. A soft iron core strengthens and radializes the field, while a spiral spring resists the coil's rotation.

3.0Important Formulas

Biot Savart Law	$d B = \frac{μ _{0}}{4 π} \cdot \frac{I d l s i n θ}{r ^{2}}$
Magnetic Field due to Thin Wire of Finite Length	$B = \frac{μ _{0} I}{4 πa} [sin ϕ_{1} + sin ϕ_{2}] inwards$
Magnetic Field due to Infinite Straight Wire	$B = \frac{μ _{0} I}{2 πa}$
Magnetic Field due to Circular Coil	$B_{0} = \frac{μ _{0} N I}{2 R}$
Magnetic Field due to Circular Arc	$B = \frac{μ _{0} I a}{4 π R}$
Magnetic Field at the Axis of Circular loop	$B_{P} = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}} = \frac{B _{0}}{( 1 + \frac{x ^{2}}{R ^{2}} ) ^{3/2}}$
Ampere’s Circuital Law	$\oint B \cdot d l = μ_{0} I_{enc}$
Magnetic Field Inside a Long Solenoid	$B = μ_{0} n I = μ_{0} (\frac{N}{l}) I$
Magnetic Field Inside the Toroid	$(R_{1} \leq r \leq R_{2}) B = μ_{0} n I n = \frac{N}{2 π r _{m}} [r_{m} = \frac{R _{1} + R _{2}}{2}] = Mean Radius of Toroid$
Magnetic Force on Moving Charge	$F = q (v \times B) = q v B sin θ$
Radius of Curvature	$R = \frac{m v}{qB} (∵ p = m v) R = \frac{p}{qB} = \frac{2 m K}{qB} K : Kinetic Energy$
Time period (T)	$T = \frac{2 πm}{qB}$
Frequency	$f = \frac{1}{T} = \frac{qB}{2 πm}$
Angular frequency	$ω = \frac{qB}{m}$
Lorentz Force	$F_{net} = q (v \times B) + (q E)$
Magnetic Force on a Current Carrying Wire	$d F = I (d l \times B)$
Magnetic Interaction Between Two Parallel Current Carrying Wires	$\frac{d F _{12}}{d l} = \frac{d F _{21}}{d l} = \frac{μ _{0} I _{1} I _{2}}{2 π d}$
Magnetic Moment	$M = N I A$
Torque on Magnetic Dipole	$τ = M \times B = MB sin θ τ = MB sin θ$
Current Sensitivity	$C.S. = \frac{Φ}{I} = \frac{N A B}{C} (rad/A)$
Voltage Sensitivity	$V.S. = \frac{Φ}{V} = \frac{Φ}{I R} = \frac{N A B}{CR} (rad/V)$
Work Done in Rotating the Coil in Uniform Magnetic Field	$W = MB (cos θ_{1} - cos θ_{2})$
Potential Energy of the Coil in Uniform Magnetic Field	$U = - MB cos θ U = - M \cdot B$
Angular SHM of magnetic dipole	$T = 2 π \frac{I}{MB} I : Moment of Inertia = \frac{M l ^{2}}{12}$
Magnetic Intensity (H)	$H = \frac{B}{μ _{0}} = A/m$
Intensity of magnetizationI	$I = \frac{M}{V} \Rightarrow A/m$
Magnetic susceptibility (m)	$χ_{m} = \frac{I}{H}$
Magnetic Permeability	$μ = \frac{B}{H} = \frac{Total magnetic field inside a material}{Magnetizing Field} = H / m$
Relative Permeabilityr	$μ_{r} = \frac{μ}{μ _{0}}$ It has no units and dimensions.
Relation between r and m	$μ_{r} = 1 + χ_{m}$

4.0Past Year Questions with Solutions on Magnetism:JEE (Mains)

Q-1.A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively, then the region of space may have :

$(A) E = 0, B = 0 (B) E = 0, B \neq = 0 (C) E \neq = 0, B = 0 (D) E \neq = 0, B \neq = 0$

Choose the most appropriate answer from the options given below :

(1)(A), (B) and (C) only

(2) (A), (C) and (D) only

(3) (A), (B) and (D) only

(4) (B), (C) and (D) only

Solution: Ans(3)

Net force on particle must be zero $i . e . q E + q (v \times B) = 0$

Possible cases are

$E$ & $B = 0$

$v \times B = 0, E = 0 q E = - q (v \times B) E \neq = 0 B \neq = 0$

Q-2.Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is 5.0 cm. The magnitude of the magnetic field at a point P midway between the wires is ____ T

(Given :0= 4 × 10-7 TmA-1)

Solution: Ans(160)

$μ T μ_{0} = 4 π \times 1 0^{- 7} TmA^{- 1} B = (\frac{μ _{0} i}{2 πa}) \times 2 = \frac{4 π \times 1 0 ^{- 7} \times 10}{π \times ( \frac{5}{2} \times 1 0 ^{- 2} )} = 16 \times 1 0^{- 5} = 160 μ T$

Q-3.A current of 200 A deflects the coil of a moving coil galvanometer through 60⁰. The current to cause deflection through 10 radian is :

$(1) 30 μ A (2) 120 μ A (3) 60 μ A (4) 180 μ A$

Solution: Ans(3)

$i \propto θ (angle of deflection) \frac{i _{2}}{i _{1}} = \frac{θ _{2}}{θ _{1}} \Rightarrow \frac{i _{2}}{200 μ A} = \frac{π /10}{π /3} = \frac{3}{10} i_{2} = 60 μ A$

Q-4.The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R1=2 m and R2=4 m carrying current I = 4A as per figure given below is × 10-7T. The value of is ______. (Centre O is common for all segments)

.The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R1=2 m and R2=4 m carrying current I = 4A as per figure given below is × 10-7 T.

Solution: Ans(3)

$R_{1} = 2 ϕ m R_{1} = 4 ϕ m α \times 1 0_{7} T \frac{μ _{0} i}{2 R _{2}} (\frac{π}{2 π}) \times + \frac{μ _{0} i}{2 R _{1}} (\frac{π}{2 π}) \times (\frac{μ _{0} i}{4 R _{2}} + \frac{μ _{0} i}{4 R _{1}}) \times \frac{4 π \times 1 0 ^{- 7} \times 4}{4 \times 4 π} + \frac{4 π \times 1 0 ^{- 7} \times 4}{4 \times 2 π}$

Q-5.The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is 1.5 × 10^–5Tm. The magnetic moment of the dipole is__________Am².

(Given : 04=10-7 TmA-1)

Solution:Ans(6)

$\frac{μ _{0}}{4 π} = 1 0^{- 7} TmA^{- 1} V = \frac{μ _{0}}{4 π} \cdot \frac{M}{r ^{2}} \Rightarrow 1.5 \times 1 0^{- 5} = 1 0^{- 7} \times \frac{M}{( 20 \times 1 0 ^{- 2} ) ^{2}} \Rightarrow M = \frac{1.5 \times 1 0 ^{- 5} \times 20 \times 20 \times 1 0 ^{- 4}}{1 0 ^{- 7}}$

Q-6.Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describe circular paths of radii R₁ and R₂ respectively. The mass ratio of X and Y is :

$(1) (\frac{R _{2}}{R _{1}})^{2} (2) (\frac{R _{1}}{R _{2}})^{2} (3) (\frac{R _{1}}{R _{2}}) (4) (\frac{R _{2}}{R _{1}})$

Solution: Ans(2)

$R = \frac{m v}{qB} = \frac{p}{qB} = \frac{2 m ( KE )}{qB} = \frac{2 m q V}{qB} R \propto m \Rightarrow m \propto R^{2} \frac{m _{1}}{m _{2}} = (\frac{R _{1}}{R _{2}})^{2}$

Q-7.A charge of $4.0 μ$ C is moving with a velocity of $4.0 \times 1 0^{6} ms^{- 1}$ along the positive y-axis under a magnetic field $B$ of strength $2 \hat{k} T$ . The force acting on the charge is $x \hat{i} N$ . The value of x is __.

Solution: Ans(32)

$q = 4 μ C, v = 4 \times 1 0^{6} \hat{j} ms^{- 1} B = 2 \hat{k} T F = q (v \times B) F = 4 \times 1 0^{- 6} (4 \times 1 0^{6} \hat{j} \times 2 \hat{k}) F = 4 \times 1 0^{- 6} \times 8 \times 1 0^{6} \hat{i} F = 32 \hat{i} N$

Q-8.The horizontal component of earth’s magnetic field at a place is $B_{H} = 3.5 \times 1 0^{- 5} T$ . A very long straight conductor carrying current of $2 A$ in the direction from South east to North West is placed. The force per unit length experienced by the conductor is ……… × $1 0^{- 6} N/m$ .

Solution: Ans(35)

$B_{H} = 3.5 \times 1 0^{- 5} T F = i lB sin θ, i = 2 A \frac{F}{l} = i B sin θ = 2 \times 3.5 \times 1 0^{- 5} \times \frac{1}{2} = 35 \times 1 0^{- 6} N/m$

Q-9.The current of 5A flows in a square loop of sides 1 m is placed in the air. The magnetic field at the centre of the loop is X $2 \times 1 0^{- 7} T$ . The value of X is_____.

Solution:Ans(40)

$B = 4 \times \frac{μ _{0} i}{4 π ( \frac{1}{2} + \frac{1}{2} )} B = 4 \times 1 0^{- 7} \times 5 \times 2 \times 2 = 402 \times 1 0^{- 7} T B = 402 \times 1 0^{- 7} T \Rightarrow χ = 40$

Q-10.A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic fieldB=B0 j as shown in figure. The magnetic force on the wire if it has a current i is :

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic fieldB=B0 j as shown in figure.

$(1) - i BR \hat{i} (2) 2 i BR \hat{j} (3) i BR \hat{j} (4) - 2 i BR \hat{j}$

Solution: Ans(4)

The magnetic force on the wire if it has a current i is

Note: Direction of magnetic field is in +k

$F = i (l \times B)$

$l = 2 R$

$F = - 2 i BR \hat{j}$

Q-11.An electron moves through a uniform magnetic field $B = B_{0} \hat{i} + 2 B_{0} \hat{j} T$ . At a particular instant of time, the velocity of electron is $u = 3 \hat{i} + 5 \hat{j} m/s$ . If the magnetic force acting on electron is $F = 5 e \hat{k} N$ , where e is the charge of an electron, then the value of $B_{0}$ is ____ T.

Solution: Ans(5)

$F = q (v \times B) 5 e \hat{k} = e (3 \hat{i} + 5 \hat{j}) \times (B_{0} \hat{i} + 2 B_{0} \hat{j}) 5 e \hat{k} = e (6 B_{0} \hat{k} - 5 B_{0} \hat{k}) B_{0} = 5 T$

Q-12. A uniform magnetic field of $2✕1 0^{- 3} T$ acts along positiveY-direction. A rectangular loop of sides 20 cm and 10 cm with a current of 5 A is the Y-Z plane. The current is in an anticlockwise sense with reference to the negative X axis. Magnitude and direction of the torque is :

$(1) 2 \times 1 0^{- 4} N - m along positive Z-direction (2) 2 \times 1 0^{- 4} N - m along negative Z-direction (3) 2 \times 1 0^{- 4} N - m along positive X-direction (3) 2 \times 1 0^{- 4} N - m along positive Y-direction$

Ans. (2)

A uniform magnetic field of 2 ✕ 10-3 T acts along positiveY-direction. A rectangular loop of sides 20 cm and 10 cm with a current of 5 A is the Y-Z plane.

$M = i A M = 5 \times (0.2) \times (0.1) (- \hat{i}) M = (0.1) (- \hat{i}) τ = M \times B = (0.1) (- \hat{i}) \times (2 \times 1 0^{- 3}) \hat{j} τ = 2 \times 1 0^{- 4} (- \hat{k}) N - m$

Q-13. Two circular coils P and Q of 100 turns each have same radius of cm. The currents in P and R are 1 A and 2 A respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is $x mT$ , where x =_________.

(Given : $μ_{0} = 4 π \times 1 0^{- 7} TmA^{- 1}$ )

Solution: Ans(20)

Q-13.Two circular coils P and Q of 100 turns each have same radius of cm. The currents in P and R are 1 A and 2 A respectively

$B_{P} = \frac{μ _{0} N _{1} i}{2 r} = \frac{μ _{0} \times 1 \times 100}{2 π} = 2 \times 1 0^{- 3} T B_{Q} = \frac{μ _{0} N _{2} i}{2 r} = \frac{μ _{0} \times 2 \times 100}{2 π} = 4 \times 1 0^{- 3} T B_{net} = B_{P}^{2} + B_{Q}^{2} = 20 mT$

Q-14.A regular polygon of 6 sides is formed by bending a wire of length $4 π$ meter. If an electric current of $4 π 3 A$ is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be $x \times 1 0^{- 7} T$ . The value of x is ____.

Solution: Ans(72)

14.A regular polygon of 6 sides is formed by bending a wire of length 4 meter.

$B = 6 (\frac{μ _{0} I}{4 π r}) (sin 3 0^{\circ} + sin 3 0^{\circ}) B = (6 \cdot \frac{1 0 ^{- 7} \times 4 π 3}{\frac{3 \times 4 π}{2 \times 6}}) = 72 \times 1 0^{- 7} T$

Q-15. A moving coil galvanometer has 100 turns and each turn has an area of 2.0 cm². The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is $x \times 1 0^{- 5}$ N-m/rad. The value of x is___.

Solution: Ans(4)

$τ = B I N A sin ϕ Cθ = B I N A sin 9 0^{\circ} C = \frac{B I N A}{θ} = \frac{0.01 \times 10 \times 1 0 ^{- 3} \times 100 \times 2 \times 1 0 ^{- 4}}{0.05} = 4 \times 1 0^{- 5} N - m / rad$

Magnetism Previous Year Questions With Solutions

1.0Introduction

Moving charges produce magnetic fields, which is the basic principle behind the magnetic effect of electric current.Oersted experiment discovered that a current-carrying wire generates a magnetic field around it, linking electricity and magnetism. According to the Right-Hand Thumb Rule, the direction of the magnetic field can be determined by pointing your thumb in the direction of current and curling your fingers. When current flows through a coil (solenoid), it creates a magnetic field like that of a bar magnet—this is called electromagnetism. Magnetic field lines form closed loops from north to south outside the magnet and never intersect. Materials like iron, cobalt, and nickel show magnetism due to aligned atomic magnetic moments. This phenomenon is widely used in motors, generators, electromagnets, and various electronic devices.

JEE Main Previous Year Solved Questions on Magnetism
JEE Adv Previous Year Solved Questions on Magnetism