Magnetic Effects Of Current Question 431
Question: Two identical long conducting wires AOB 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 T’ 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:
Options:
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}\times I_{2}^{2} )}^{1/2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Net magnetic field, $ B=\sqrt{B_{1}^{2}+B_{2}^{2}} $ $ =\sqrt{{{( \frac{{\mu_{0}}I_{1}}{2\pi d} )}^{2}}+{{( \frac{{\mu_{0}}I_{2}}{2\pi d} )}^{2}}}=\frac{{\mu_{0}}}{2\pi d}\sqrt{I_{1}^{2}+I_{2}^{2}} $
BETA
