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

A current carrying circular loop of radius 2 cm with unit normal $\hat{n}=\frac{\hat{k}+\hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\vec{B}=B_o(3 \hat{i}+2 \hat{k})$. If $B_o=4 \times 10^{-3} \mathrm{~T}$ and current $I=100 \sqrt{2} \mathrm{~A}$, the torque experienced by the loop is $\_\_\_\_$ Wb.A. ( $\pi=3.14$ )

A

$$ 16 \times 10^{-5} \hat{k} $$

B

$$ 5024 \times 10^{-7} \hat{k} $$

C

$$ 5024 \times 10^{-7} \hat{i} $$

D

$$ 5024 \times 10^{-7} \hat{j} $$

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

A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm . The magnetic field at the mid point between the two wires is $\_\_\_\_$ $\mu \mathrm{T}$.

$$ \left(\frac{\mu_{\mathrm{o}}}{4 \pi}=10^{-7} \mathrm{~N} / \mathrm{A}^2\right) $$

A

30

B

300

C

150

D

0.0

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

A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A . The magnetic energy stored inside the cube is $\alpha \times 10^{-14} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ .

$$ \left(\mu_{\mathrm{o}}=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}, \pi=3.14\right) $$

A

6.28

B

$$6. 28 \times 10^{-6} $$

C

628

D

$$ 6 .28 \times 10^{-4} $$

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

A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v}=\left(\frac{v_0}{\sqrt{2}} \hat{x}+\frac{v_0}{\sqrt{2}} \hat{y}\right)$. There exists a uniform magnetic field $\vec{B}=B_0 \hat{z}$ and a space varying electric field $\vec{E}=E_{\mathrm{o}} \mathrm{e}^{-\lambda x} \hat{x}$ within the region $0 \leqslant x \leqslant L$. After travelling a distance such that $x$-coordinate has changed from $x=0$ to $x=L$, the change in the kinetic energy is $\_\_\_\_$ .

A

$\frac{q E_0}{\lambda}\left[1-e^{-\lambda L}\right]$

B

$\left(\frac{v_0 q B_0}{2 \lambda}\right)\left[2-e^{-2 \lambda L}\right]$

C

$$ \frac{q E_0}{\lambda}\left[1+e^{-\lambda L}\right] $$

D

$$ q\left(\frac{E_0+v_0 B_0}{\lambda}\right)\left[1-e^{-\lambda L / 2}\right] $$

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