question_answer

A wire carrying current \[I\]  has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to x-axis while semicircular portion of radius B. is lying in Y–Z plane. Magnetic field at point O is

[NEET  2015]

A)  \[\mathbf{B}=\frac{{{\mu }_{0}}}{4\pi }\frac{l}{R}(\pi \hat{i}+2\hat{k})\]

B)       \[\mathbf{B}=-\frac{{{\mu }_{0}}}{4\pi }\frac{l}{R}(\pi \hat{i}-2\hat{k})\]

C)  \[\mathbf{B}=-\frac{{{\mu }_{0}}}{4\pi }\frac{l}{R}(\pi \hat{i}+2\hat{k})\]

D)  \[\mathbf{B}=\frac{{{\mu }_{0}}}{4\pi }\frac{l}{R}(\pi \hat{i}-2\hat{k})\]

Correct Answer: C

Solution :

[c] The magnetic field in the different regions is given by

\[{{\mathbf{B}}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{I}{R}(-\mathbf{\hat{k}})\]

\[{{\mathbf{B}}_{3}}=\frac{{{\mu }_{0}}I}{4\pi R}(-\mathbf{\hat{k}})\]

\[{{\mathbf{B}}_{2}}=\frac{{{\mu }_{0}}I}{4\pi R}(-\mathbf{\hat{i}})\]

The net magnetic field at the centre O is

\[\mathbf{B}={{\mathbf{B}}_{\mathbf{1}}}+{{\mathbf{B}}_{\mathbf{2}}}+{{\mathbf{B}}_{\mathbf{3}}}\]

\[=\frac{{{\mu }_{0}}I}{4\pi R}(-2\hat{k})+\frac{{{\mu }_{0}}I}{4R}(-\hat{i})\]

\[=-\frac{{{\mu }_{o}}I}{4\pi R}(2\hat{k}+\pi \hat{i})\]