### NEET Solved Paper 2014 Question 31

##### 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 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 [AIPMT 2014]

#### Options:

A) $ \frac{{\mu_0}}{2\pi d}( \frac{l _1}{l _2} ) $

B) $ \frac{{\mu_0}}{2\pi d}(l _1+l _2) $

C) $ \frac{{\mu_0}}{2\pi d}(l_1^{2}+l_2^{2}) $

D) $ \frac{{\mu_0}}{2\pi d}{{(l_1^{2}+l_2^{2})}^{1/2}} $

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

The point P is lying at a distance d along the z axis.

$ | {{\mathbf{B}}_1} |=\frac{{\mu_0}}{2\pi }\frac{l _1}{d} $ and

$ | {{\mathbf{B}}_2} |=\frac{{\mu_0}}{2\pi }\frac{l _2}{d} $ $ B _{net}=\sqrt{B_1^{2}+B_2^{2}}, $ $ B _{net}=\frac{{\mu_0}}{2\pi }\frac{1}{d}{{(l_1^{2}+l_2^{2})}^{1/2}} $