JEE Main 2026 — Physics Electromagnetism

The magnetic dipole moment of the current carrying loop is given by $\vec{M} = I A \hat{n}$M=IAn^.

Given radius $r = 2$r=2 cm $= 2 \times 10^{-2}$=2×10−2 m, the area of the loop is:

$A = \pi r^2 = \pi (2 \times 10^{-2})^2 = 4\pi \times 10^{-4}$A=πr2=π(2×10−2)2=4π×10−4 m$^2$2

Substituting the given values $I = 100\sqrt{2}$I=1002​ A and $\hat{n} = \dfrac{\hat{i}+\hat{k}}{\sqrt{2}}$n^=2​i^+k^​:

$$\vec{M} = (100\sqrt{2}) \times (4\pi \times 10^{-4}) \times \left(\dfrac{\hat{i}+\hat{k}}{\sqrt{2}}\right)$$M=(1002​)×(4π×10−4)×(2​i^+k^​)

$\vec{M} = 4\pi \times 10^{-2} (\hat{i}+\hat{k})$M=4π×10−2(i^+k^) A$\cdot$⋅m$^2$2

The torque experienced by the loop in the magnetic field is $\vec{\tau} = \vec{M} \times \vec{B}$τ=M×B.

Given $\vec{B} = 4 \times 10^{-3} (3\hat{i}+2\hat{k})$B=4×10−3(3i^+2k^) T, we have:

$$\vec{\tau} = [4\pi \times 10^{-2} (\hat{i}+\hat{k})] \times [4 \times 10^{-3} (3\hat{i}+2\hat{k})]$$τ=[4π×10−2(i^+k^)]×[4×10−3(3i^+2k^)]

$$\vec{\tau} = 16\pi \times 10^{-5} [(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k})]$$τ=16π×10−5[(i^+k^)×(3i^+2k^)]

Evaluating the cross product:

$$(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k}) = (\hat{i} \times 3\hat{i}) + (\hat{i} \times 2\hat{k}) + (\hat{k} \times 3\hat{i}) + (\hat{k} \times 2\hat{k})$$(i^+k^)×(3i^+2k^)=(i^×3i^)+(i^×2k^)+(k^×3i^)+(k^×2k^)

$$= 0 + 2(-\hat{j}) + 3(\hat{j}) + 0 = \hat{j}$$=0+2(−j^​)+3(j^​)+0=j^​

Therefore, the torque is:

$$\vec{\tau} = 16\pi \times 10^{-5} \hat{j}$$τ=16π×10−5j^​

Substituting $\pi = 3.14$π=3.14:

$\vec{\tau} = 16 \times 3.14 \times 10^{-5} \hat{j} = 50.24 \times 10^{-5} \hat{j} = 5024 \times 10^{-7} \hat{j}$τ=16×3.14×10−5j^​=50.24×10−5j^​=5024×10−7j^​ Wb$\cdot$⋅A