Q149 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Physics
Two infinite wires carrying opposite electrical currents $I$ and $i$ are placed a distance $x$ apart.
A point $P$ at a distance $y$ away from the wire carrying current $i$ is shown in the figure. If the magnetic field is zero at point $P$, then the magnitude of $i$ is


- A$l\left(\frac{x}{x+y}\right)$
- B$\prime\left(\frac{2 x}{x+y}\right)$
- Correct answerC$I\left(\frac{y}{x+y}\right)$
- D$\left(\frac{2 y}{x+y}\right)$
Solution
The given situation is shown in the following figure

Magnetic field at point $P$ due to the wire having current $I$ is given as $B_I=\frac{\mu_0}{2 \pi} \cdot \frac{I}{x+y}$ (downward) Magnetic field at point $P$ due to the wire having current $i$ is given as $B_i=\frac{\mu_0}{2 \pi} \cdot \frac{i}{y}$ (upward) According to question, net magnetic field at point $P$ is zero, hence $\begin{gathered} B_I=B_i \\ \Rightarrow \frac{\mu_0}{2 \pi} \cdot \frac{I}{x+y}=\frac{\mu_0}{2 \pi} \cdot \frac{i}{y} \Rightarrow \frac{I}{x+y}=\frac{i}{y} \Rightarrow i=I\left(\frac{y}{x+y}\right) \end{gathered}$

Magnetic field at point $P$ due to the wire having current $I$ is given as $B_I=\frac{\mu_0}{2 \pi} \cdot \frac{I}{x+y}$ (downward) Magnetic field at point $P$ due to the wire having current $i$ is given as $B_i=\frac{\mu_0}{2 \pi} \cdot \frac{i}{y}$ (upward) According to question, net magnetic field at point $P$ is zero, hence $\begin{gathered} B_I=B_i \\ \Rightarrow \frac{\mu_0}{2 \pi} \cdot \frac{I}{x+y}=\frac{\mu_0}{2 \pi} \cdot \frac{i}{y} \Rightarrow \frac{I}{x+y}=\frac{i}{y} \Rightarrow i=I\left(\frac{y}{x+y}\right) \end{gathered}$
Hence, the correct answer is option: 'C'.