Question

Two infinitely long straight wires lie in the $xy$-plane along the lines $x = \pm R$. The wire located at $x = + R$ carries a constant current $I_1$ and the wire located at $x = - R $ carries a constant current $I_2$. A circular loop of radius

• A. If $I_1 = I_2$, then $\vec{B}$ cannot be equal to zero at the origin (0, 0, 0)
• B. If $I_1 > 0 $ and $I_2 < 0 $, then $\vec{B}$ can be equal to zero at the origin (0, 0, 0)
• C. If $I_1 < 0 $ and $I_2 > 0 $, then $\vec{B}$ can be equal to zero at the origin (0, 0, 0)
• D. If $I_1 = I_2$, then the z-component of the magnetic field at the centre of the loop is $( - \frac{\mu_0 I}{2R})$

Answer 1

The Correct Option is D

(A) At origin, $\vec{B} = 0 $ due to two wires if $I_1 = I_2$ , hence $(\vec{B}_{net})$ at origin is equal to $\vec{B}$ due to ring, which is non-zero.
(B) If $I_1 > 0$ and $I_2 < 0, \vec{B}$ at origin due to wires will be along $+ \hat{k}$ direction and $\vec{B}$ due to ring is along $- \hat{k}$ direction and hence $\vec{B}$ can be zero at origin.
(C) If $I_1 < 0$ and $I_2 > 0, \vec{B}$ at origin due to wires is along $- \hat{k}$ and also along $ - \hat{k}$ due to ring $-\hat{k}$ and also along $- \hat{k}$ due to ring, hence $\vec{B}$ cannot be zero.
(D) ....
At centre of ring, $\vec{B}$ due to wires is along x -axis,
hence z-component is only because of ring which $\vec{B} = \frac{\mu_0 i }{2R} ( - \hat{k})$