JEE Physics 2024 Question with Solution

Given: Two insulated circular loops A and B of radius $a$ each carry current $I$ and are perpendicular to each other.

Find: The magnitude of the net magnetic induction at the common center.

For one circular loop, the magnetic field at the center is

$$B = \frac{\mu_0 I}{2a}$$

The direction of this field is perpendicular to the plane of the loop by the right-hand thumb rule.

For loop A,

$$B_A = \frac{\mu_0 I}{2a}$$

For loop B,

$$B_B = \frac{\mu_0 I}{2a}$$

Since the two loops are perpendicular, the magnetic field vectors at the center are also perpendicular to each other.

Using vector addition,

$$|\vec{B}_{\text{net}}| = \sqrt{B_A^2 + B_B^2}$$

Substituting,

$$|\vec{B}_{\text{net}}| = \sqrt{\left(\frac{\mu_0 I}{2a}\right)^2 + \left(\frac{\mu_0 I}{2a}\right)^2}$$ $$|\vec{B}_{\text{net}}| = \sqrt{2\left(\frac{\mu_0 I}{2a}\right)^2}$$ $$|\vec{B}_{\text{net}}| = \frac{\sqrt{2}\mu_0 I}{2a} = \frac{\mu_0 I}{\sqrt{2}a}$$

Therefore, the magnitude of the magnetic induction at the center is $\frac{\sqrt{2}\mu_0 I}{2a}$.

The solution working does not match the listed option C. Based on the extracted source, option A is also not equivalent to this value as written, and the printed answer key appears inconsistent with the worked solution.

Given: Each loop produces magnetic field of magnitude $\frac{\mu_0 I}{2a}$ at the center.

Find: Resultant magnetic field when the two fields are perpendicular.

Let the field due to one loop be

$$B = \frac{\mu_0 I}{2a}$$

Because the loops are in mutually perpendicular planes, their magnetic fields at the center are mutually perpendicular.

So,

$$B_{\text{net}} = \sqrt{B^2 + B^2}$$ $$B_{\text{net}} = \sqrt{2B^2} = \sqrt{2}\,B$$ $$B_{\text{net}} = \sqrt{2}\left(\frac{\mu_0 I}{2a}\right)$$ $$B_{\text{net}} = \frac{\mu_0 I}{\sqrt{2}a}$$

Thus, the correct magnitude from the solution is $\frac{\mu_0 I}{\sqrt{2}a}$.