1
JEE Main 2026 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A 30 cm long solenoid has 10 turns per cm and area of $5 \mathrm{~cm}^2$. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s . The e.m.f. induced in the coil is $\alpha \times 10^{-5} \mathrm{~V}$. The value $\alpha$ is $\_\_\_\_$ .

A

60

B

12

C

120

D

34

2
JEE Main 2026 (Online) 6th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as $B=0.4 \sin (300 t)$ Tesla. The normal to the plane of loop makes an angle of $60^{\circ}$ with the field. The maximum induced emf produced in the loop is $\_\_\_\_$ mV .

A

12

B

18

C

21

D

24

3
JEE Main 2026 (Online) 6th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An inductor of inductance 10 mH having resistance of $100 \Omega$ is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is 2 mA and 4 mA is $\_\_\_\_$ .

JEE Main 2026 (Online) 6th April Morning Shift Physics - Electromagnetic Induction Question 4 English
A

4/3

B

3/4

C

5/3

D

3/5

4
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(\mathrm{r})=B_{\mathrm{o}} \mathrm{e}^{-\lambda r} ; \lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

A

$$ B_o \omega\left[\frac{1}{\lambda^2}-e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

B

$$ B_o \omega\left[\frac{1}{\lambda^2}+e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

C

$$ B_o \omega\left[\frac{4}{\lambda^2}-e^{-2 \lambda L}\left(\frac{1}{\lambda^2}+\frac{2 L}{\lambda}\right)\right] $$

D

$$ B_0 \omega\left[\frac{3}{\lambda^2}-e^{-3 \lambda L}\left(\frac{3}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

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