Q67 / 90 · JEE Main 2010 · Physics
Two long parallel wires are at a distance $2 \mathrm{~d}$ apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line $\mathrm{XX}$ ' is given by
- Correct answerA
- B
- C
- D
Solution
The magnetic field in between because of each will be in opposite direction
$$
\begin{aligned}
B_{\text {in between }} & =\frac{\mu_0 i}{2 \pi x} \hat{j}-\frac{\mu_0 i}{2 \pi(2 d-x)}(-\hat{j}) \\
& =\frac{\mu_0 i}{2 \pi}\left[\frac{1}{x}-\frac{1}{2 d-x}\right](\hat{j})
\end{aligned}
$$
at $\quad x=d, \quad B_{\text {in between }}=0$
for $\quad x < d, \quad B_{\text {in between }}=(\hat{j})$
for $\quad x>d, \quad B_{\text {in between }}=(-\hat{j})$
towards $\mathrm{x}$ net magnetic field will add up and direction will be $(-\hat{\mathrm{j}})$
towards $\mathrm{x}^{\prime}$ net magnetic field will add up and direction will be $(\hat{\mathrm{j}})$
Hence, the correct answer is option: 'A'.