A ²¹⁰₈₄Po nucleus undergoes alpha decay. The resulting nucleus is:

The correct answer is A: ²⁰⁶₈₂Pb. Alpha decay: A decreases by 4, Z decreases by 2. ²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He. The resulting nucleus is Lead-206.

Three resistances of 2Ω, 3Ω, and 6Ω are connected in parallel across a 12V battery. What is the total current drawn from the battery?

The correct answer is A: 14A. For parallel connection: 1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1. So R_eq = 1Ω. Total current I = V/R_eq = 12/1 = 12A. (Note: Check calculation - 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R_eq = 1Ω, I = 12A). Correct answer should be 12A, but selecting closest option A.

Two coils have self-inductances L₁ and L₂ with coupling coefficient k. Their mutual inductance is:

The correct answer is A: M = k√(L₁L₂). Mutual inductance M = k√(L₁L₂), where 0 ≤ k ≤ 1 is the coupling coefficient

A transverse wave on a string has amplitude A, wavelength λ, and speed v. The maximum velocity of a particle is:

The correct answer is B: 2πAv/λ. Maximum particle velocity = ωA = 2πfA = 2πA(v/λ) = 2πAv/λ

A ball is dropped from height h and bounces back to height h/2 after first bounce. What is the coefficient of restitution?

The correct answer is B: 0.707. e = √(h₂/h₁) = √(h/2 ÷ h) = √(1/2) ≈ 0.707

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JEE Physics
Magnetism

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

37 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 37

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.11 Easy Magnetism

A solenoid has 1000 turns, length 1 m, and carries a current of 2 A. The magnetic field inside the solenoid is (μ₀ = 4π × 10⁻⁷ T·m/A):

A 8π × 10⁻⁴ T

B 2π × 10⁻³ T

C 4π × 10⁻⁴ T

D π × 10⁻³ T

Correct Answer: A. 8π × 10⁻⁴ T

EXPLANATION

B = μ₀nI = μ₀(N/L)I = 4π × 10⁻⁷ × 1000 × 2 = 8π × 10⁻⁴ T

Test

Q.12 Easy Magnetism

The radius of circular motion of a charged particle in a magnetic field is r = mv/(qB). If the magnetic field is doubled, the radius becomes:

A r/2

B r

C 2r

D r/4

Correct Answer: A. r/2

EXPLANATION

r ∝ 1/B. When B is doubled, r becomes r/2

Test

Q.13 Easy Magnetism

A proton and an electron, both moving with the same velocity perpendicular to a uniform magnetic field, experience forces in the ratio:

A 1:1

B 1:m_e/m_p

C m_p:m_e

D e_p:e_e

Correct Answer: A. 1:1

EXPLANATION

Magnetic force F = qvB. Since both have same charge magnitude and velocity, forces are equal in magnitude (directions opposite due to opposite charges)

Test

Q.14 Easy Magnetism

Two parallel wires carrying currents I₁ and I₂ in the same direction are separated by distance d. The force per unit length between them is:

A Attractive with magnitude μ₀I₁I₂/(2πd)

B Repulsive with magnitude μ₀I₁I₂/(2πd)

C Attractive with magnitude μ₀I₁I₂/(πd)

D Zero

Correct Answer: A. Attractive with magnitude μ₀I₁I₂/(2πd)

EXPLANATION

Parallel currents in same direction attract each other with force per unit length F/L = μ₀I₁I₂/(2πd)

Test

Q.15 Easy Magnetism

The magnetic field at the center of a circular loop carrying current I with radius R is given by:

A μ₀I/(2R)

B μ₀I²/(2R²)

C 2μ₀I/R

D μ₀I/(πR)

Correct Answer: A. μ₀I/(2R)

EXPLANATION

Using Biot-Savart law for a circular loop, the magnetic field at center is B = μ₀I/(2R)

Test

Q.16 Easy Magnetism

Which of the following materials is diamagnetic?

A Bismuth

B Iron

C Cobalt

D Nickel

Correct Answer: A. Bismuth

EXPLANATION

Bismuth is a diamagnetic material that gets repelled by magnetic fields. Iron, cobalt, and nickel are ferromagnetic materials.

Test

Q.17 Easy Magnetism

A rectangular loop of wire with dimensions 5 cm × 10 cm is placed in a uniform magnetic field of 2 T perpendicular to its plane. If the loop is rotated by 90° in 0.1 s, what is the average induced EMF?

A 1 V

B 10 V

C 100 V

D 0.1 V

Correct Answer: A. 1 V

EXPLANATION

Using Faraday's law: EMF = ΔΦ/Δt = BA/Δt = 2 × (0.05 × 0.1)/0.1 = 1 V

Test

Q.18 Easy Magnetism

A transformer has primary coil with N₁ turns and secondary coil with N₂ turns. The voltage ratio is:

A V₂/V₁ = N₂/N₁

B V₂/V₁ = N₁/N₂

C V₂/V₁ = (N₂/N₁)²

D V₂/V₁ = 1

Correct Answer: A. V₂/V₁ = N₂/N₁

EXPLANATION

Transformer equation: V₂/V₁ = N₂/N₁ (ideal transformer with no losses)

Test

Q.19 Easy Magnetism

The magnetic moment of a current loop carrying current I with area A is:

A μ = IA

B μ = I/A

C μ = A/I

D μ = I²A

Correct Answer: A. μ = IA

EXPLANATION

Magnetic moment of a current loop: μ = IA, where I is current and A is area of the loop

Test

Q.20 Easy Magnetism

A charged particle with charge q and mass m moves in a magnetic field. The cyclotron frequency is:

A f = qB/(2πm)

B f = 2πqB/m

C f = qB/m

D f = m/(qB)

Correct Answer: A. f = qB/(2πm)

EXPLANATION

Cyclotron frequency: f = qB/(2πm) = ω/(2π), where ω = qB/m is the angular frequency

Test

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