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 :
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 \[\mathbf{B}={{\mathbf{B}}_{\mathbf{1}}}+{{\mathbf{B}}_{\mathbf{2}}}+{{\mathbf{B}}_{\mathbf{3}}}\] \[=\frac{{{\mu }_{0}}I}{4\pi R}(-2\hat{k})+\frac{{{\mu }_{0}}I}{4R}(-\hat{i})=-\frac{{{\mu }_{o}}I}{4\pi R}(2\hat{k}+\pi \hat{i})\]You need to login to perform this action.
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