JEE Advanced 2018 · Paper 1, Physics question 5

Wires at the origin. Each wire is a distance $R$ from the origin. Using $d\vec l\times\hat r$:

• wire at $x=+R$ (current $I_1$ along $+\hat j$): $+\dfrac{\mu_0I_1}{2\pi R}\hat k$;
• wire at $x=-R$ (current $I_2$ along $+\hat j$): $-\dfrac{\mu_0I_2}{2\pi R}\hat k$.

Net: $\dfrac{\mu_0(I_1-I_2)}{2\pi R}\hat k$.

Loop at the origin. The origin is on the loop's axis at distance $\sqrt3R$. A clockwise current (seen from above) gives a field along $-\hat k$:
$$B_{loop}=\frac{\mu_0IR^2}{2(R^2+3R^2)^{3/2}}=\frac{\mu_0I}{16R}.$$

Cancellation needs $\dfrac{I_1-I_2}{2\pi}=\dfrac I{16}$, i.e. $I_1-I_2>0$.

• (A) $I_1=I_2$: the wires cancel and the loop's field remains. $\vec B\ne0$. True.
• (B) $I_1>0,\ I_2<0$: $I_1-I_2>0$, so cancellation is possible. True.
• (C) $I_1<0,\ I_2>0$: $I_1-I_2<0$, both fields point along $-\hat k$. False.
• (D) At the loop's centre each wire is $2R$ away. By symmetry, when $I_1=I_2$ the $z$-components of their fields there are equal and opposite. Only the loop contributes: $B_z=-\dfrac{\mu_0I}{2R}$. True.

(A), (B), (D)