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JEE Physics
Magnetism

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

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Topics in JEE Physics
Q.1 Hard Magnetism
A solenoid with N turns, length L, and cross-sectional area A is wound with wire of resistance R. When connected to a voltage source V, the magnetic energy stored is:
A V²NL/(2R)
B V²μ₀N²A/(2RL)
C V²μ₀N²A/(2R²L)
D V²L/(2Rμ₀N²A)
Correct Answer:  C. V²μ₀N²A/(2R²L)
EXPLANATION

Current I = V/R. Self-inductance L = μ₀N²A/L. Magnetic energy = LI²/2 = (μ₀N²A/L)·(V²/R²)/2 = V²μ₀N²A/(2R²L)

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Q.2 Hard Magnetism
A charged particle enters a region with perpendicular electric and magnetic fields with velocity v. For the particle to pass undeflected, the condition is:
A E = vB
B E = B/v
C E = v/B
D E = B²/v
Correct Answer:  A. E = vB
EXPLANATION

For undeflected motion, electric force equals magnetic force: qE = qvB, which gives E = vB. This is the principle of a velocity selector.

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Q.3 Hard Magnetism
A rectangular loop of dimensions a × b is placed in a non-uniform magnetic field where B varies as B = B₀(1 + kx), where x is the distance from a reference line. The net force on the loop is:
A Zero
B B₀kbIa
C B₀kIab
D 2B₀kIa
Correct Answer:  C. B₀kIab
EXPLANATION

In a non-uniform field, the forces on opposite sides of the loop are unequal. The net force depends on the field gradient. F = I·∫(dB/dx)·dA = I·b·∫B₀k·da = B₀kIab (approximately, for small variations).

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Q.4 Hard Magnetism
A proton and an alpha particle (He²⁺ nucleus) are accelerated through the same potential difference. They are then made to move perpendicular to a uniform magnetic field. The ratio of their radii of curvature is:
A 1:1
B 1:2
C 2:1
D 1:4
Correct Answer:  C. 2:1
EXPLANATION

From qVB = mv²/2 and r = mv/(qB), we get r = √(2mV/q)/B. For proton (m=m_p, q=e) and alpha (m=4m_p, q=2e): r_p/r_α = √(m_p/(4m_p))·√(2e/e) = √(1/4)·√2 = √(1/2)·√2 = 2/1

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Q.5 Medium Magnetism
The SI unit of magnetic flux density (magnetic field) is:
A Weber (Wb)
B Tesla (T)
C Gauss (G)
D Henry (H)
Correct Answer:  B. Tesla (T)
EXPLANATION

The SI unit of magnetic field (flux density) is Tesla (T). 1 T = 1 Wb/m² = 1 kg/(A·s²). Weber is the unit of magnetic flux, Gauss is CGS unit, and Henry is unit of inductance.

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Q.6 Medium Magnetism
A long straight wire carrying current I produces a magnetic field at distance r. If the current is doubled and distance is halved, the magnetic field becomes:
A 2 times
B 4 times
C 8 times
D Same as before
Correct Answer:  B. 4 times
EXPLANATION

B = μ₀I/(2πr). If I → 2I and r → r/2, then B_new = μ₀(2I)/(2π(r/2)) = 4·μ₀I/(2πr) = 4B_initial

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Q.7 Medium Magnetism
The permeability of free space μ₀ has the value:
A 8.85 × 10⁻¹² F/m
B 4π × 10⁻⁷ T·m/A
C 3 × 10⁸ m/s
D 6.63 × 10⁻³⁴ J·s
Correct Answer:  B. 4π × 10⁻⁷ T·m/A
EXPLANATION

The permeability of free space μ₀ = 4π × 10⁻⁷ T·m/A. Option A is permittivity ε₀, option C is speed of light, and option D is Planck's constant.

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Q.8 Medium Magnetism
A charged particle with charge q and mass m is moving with speed v in a circular path of radius r in a magnetic field. The magnetic field strength is:
A B = mv/(qr)
B B = qr/(mv)
C B = qv/(mr)
D B = mr/(qv)
Correct Answer:  A. B = mv/(qr)
EXPLANATION

From qvB = mv²/r (centripetal force equals magnetic force), we get B = mv/(qr). This is the relationship between field strength, particle properties, and circular path radius.

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Q.9 Medium Magnetism
The magnetic susceptibility of a paramagnetic material is:
A Negative and large in magnitude
B Positive and small
C Zero
D Negative and small
Correct Answer:  B. Positive and small
EXPLANATION

Paramagnetic materials have positive but small magnetic susceptibility (χ > 0, typically 10⁻⁵ to 10⁻³). Diamagnetic materials have small negative susceptibility, and ferromagnetic materials have large positive susceptibility.

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Q.10 Medium Magnetism
A conducting rod of length L moves with velocity v perpendicular to its length in a magnetic field B. The motional EMF induced is maximum when:
A v is parallel to B
B v is perpendicular to B
C v is at 45° to B
D B is zero
Correct Answer:  B. v is perpendicular to B
EXPLANATION

Motional EMF = B·L·v·sinθ, where θ is the angle between v and B. EMF is maximum when sinθ = 1, i.e., when v is perpendicular to B.

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