Moving Charges and Magnetism - Result Question 35
36. Two identical long conducting wires $A O B$ and COD are placed at right angle to each other, with one above other such that ’ $O$ ’ is their common point for the two. The wires carry $I_1$ and $I_2$ currents respectively. Point ’ $P$ ’ is lying at distance ’ $d$ ’ from ’ $O$ ’ along a direction perpendicular to the plane containing the wires. The magnetic field at the point ’ $P$ ’ will be :
[2014]
(a) $\frac{\mu_0}{2 \pi d}(\frac{I_1}{I_2})$
(b) $\frac{\mu_0}{2 \pi d}(I_1+I_2)$
(c) $\frac{\mu_0}{2 \pi d}(I_1^{2}-I_2^{2})$
(d) $\frac{\mu_0}{2 \pi d}(I_1^{2}+I_2^{2})^{1 / 2}$
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Answer:
Correct Answer: 36. (d)
Solution:
- (d) Net magnetic field, $B=\sqrt{B_1^{2}+B_2^{2}}$
$ \begin{aligned} & =\sqrt{(\frac{\mu_0 I_1}{2 \pi d})^{2}+(\frac{\mu_0 I_2}{2 \pi d})^{2}} \\ & \quad(\because B_1=\frac{\mu_0 I_1}{2 \pi d} \text{ and } B_2=\frac{\mu_0 I_2}{2 \pi d}) \\ & =\frac{\mu_0}{2 \pi d} \sqrt{I_1^{2}+I_2^{2}} \end{aligned} $
