Q115 / 125 · BITSAT 2013 · Physics
Two parallel conductors carry current in opposite directions as shown in figure. One conductor carries a current of $10.0 \mathrm{~A}$. Point $\mathrm{C}$ is a distance
$\frac{\mathrm{d}}{2}$ to the right of the $10.0 \mathrm{~A}$ current. If $\mathrm{d}=18 \mathrm{~cm}$ and I is adjusted so that the magnetic field at $\mathrm{Cis}$ zero, the value of the current $I$ is
- A$10.0 \mathrm{~A}$
- Correct answerB$30.0 \mathrm{~A}$
- C$8.0 \mathrm{~A}$
- D$18.0 \mathrm{~A}$ #
Solution
The magnetic field at $\mathrm{C}$ due to first conductor
is $\mathrm{B}_{1}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{3 \mathrm{~d} / 2}$ (since, point $\mathrm{C}$ is separated
by $\mathrm{d}+\frac{\mathrm{d}}{2}=\frac{3 \mathrm{~d}}{2}$ from 1 st conductor). The direction of field is perpendicular to the plane of paper and directed outwards. The magnetic field at $\mathrm{C}$ due to second conductor is $\mathrm{B}_{2}=\frac{\mu_{0}}{2 \pi} \frac{10}{\mathrm{~d} / 2}$ (since, point $\mathrm{C}$
is separated by $\frac{\mathrm{d}}{2}$ from 2 nd conductor) The direction of field is perpendicular to the plane of paper and directed inwards. Since, direction of $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ at point $\mathrm{C}$ is in opposite direction and the magnetic field at C is zero, therefore,
$$
\begin{array}{c}
\mathrm{B}_{1}=\mathrm{B}_{2} \\
\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{3 \mathrm{~d} / 2}=\frac{\mu_{0}}{2 \pi} \frac{10}{\mathrm{~d} / 2}
\end{array}
$$
On solving $\mathrm{I}=30.0 \mathrm{~A}$
Hence, the correct answer is option: 'B'.