JEE Physics 2025 Question with Solution

Given: $I = 12 \, A$, $R_1 = 6\pi \, m$, $R_2 = 4\pi \, m$, and $\mu_0 = 4\pi \times 10^{-7} \, T \, m \, A^{-1}$.

Find: The value of $k$ if the resultant magnetic field at $O$ is $k \times 10^{-7} \, T$.

The magnetic field at the center due to a semicircular arc is

$$B = \frac{\mu_0 I}{4R}$$

The straight segments passing through the line of the center produce zero magnetic field at $O$.

So, only the two semicircular segments contribute. Their magnetic fields are in opposite directions, hence

$$B_{\text{net}} = \left| \frac{\mu_0 I}{4R_2} - \frac{\mu_0 I}{4R_1} \right|$$

Substituting the given values,

$$B_{\text{net}} = \frac{4\pi \times 10^{-7} \times 12}{4} \left( \frac{1}{4\pi} - \frac{1}{6\pi} \right)$$ $$B_{\text{net}} = 12\pi \times 10^{-7} \left( \frac{3 - 2}{12\pi} \right)$$ $$B_{\text{net}} = 12\pi \times 10^{-7} \left( \frac{1}{12\pi} \right)$$ $$B_{\text{net}} = 1 \times 10^{-7} \, T$$

Comparing with $k \times 10^{-7} \, T$, we get

$$k = 1$$

Therefore, the value of $k$ is $1$.

Given: Two semicircular arcs of radii $R_1$ and $R_2$ carry the same current $I$ in opposite senses about the center $O$.

Find: The value of $k$.

Since the field due to a semicircle is inversely proportional to radius,

$$B \propto \frac{1}{R}$$

So the resultant field is directly

$$B_0 = \left|B_{R_1} - B_{R_2}\right| = \left| \frac{\mu_0 I}{4R_2} - \frac{\mu_0 I}{4R_1} \right|$$

Now substitute:

$$B_0 = \frac{4\pi \times 10^{-7} \times 12}{4} \left| \frac{1}{4\pi} - \frac{1}{6\pi} \right|$$ $$B_0 = 12\pi \times 10^{-7} \left| \frac{1}{12\pi} \right| = 1 \times 10^{-7} \, T$$

Hence,

$$k = 1$$

The required answer is $1$.