JEE Main 2025 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

A coil of area A and N tums is rotating with angular velocity to in a uniform magnetic field $\vec{B}$ about an axis perpendicular to $\vec{B}$. Magnetic flux $\varphi$ and induced emf $\varepsilon$ across it, at an instant when $\vec{B}$ is parallel to the plane of coil, are :

φ = AB, φ = NABω

φ = AB, φ = 0

φ = 0, ε = 0

φ = 0, ε = NABω

JEE Main 2025 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

Consider I₁ and I₂ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L₁ = self inductance of coil 1, M₁₂ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :

e₁ = -L₁$\frac{dI_2}{dt}$ - M₁₂$\frac{dI_1}{dt}$

e₁ = -L₁$\frac{dI_1}{dt}$ + M₁₂$\frac{dI_2}{dt}$

e₁ = -L₁$\frac{dI_1}{dt}$ - M₁₂$\frac{dI_1}{dt}$

e₁ = -L₁$\frac{dI_1}{dt}$ - M₁₂$\frac{dI_2}{dt}$

JEE Main 2025 (Online) 28th January Evening Shift

MCQ (Single Correct Answer)

-1

A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10 $ \pi $ rad s^-1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? $(\pi=3.14)$

0.5024 V

0.0628 V

0.2512 V

0.1256 V

JEE Main 2025 (Online) 23rd January Morning Shift

MCQ (Single Correct Answer)

-1

Regarding self-inductance:

A. The self-inductance of the coil depends on its geometry.

B. Self-inductance does not depend on the permeability of the medium.

C. Self-induced e.m.f. opposes any change in the current in a circuit.

D. Self-inductance is electromagnetic analogue of mass in mechanics.

E. Work needs to be done against self-induced e.m.f. in establishing the current.

Choose the correct answer from the options given below:

A, B, C, E only

A, B, C, D only

A, C, D, E only

B, C, D, E only

JEE Main Subjects