A) \[\frac{{{\mu }_{0}}I}{4a}\]
B) \[\frac{{{\mu }_{0}}I}{4\pi a}\sqrt{{{\pi }^{2}}+4}\]
C) \[\frac{{{\mu }_{0}}I}{4a}+\frac{{{\mu }_{0}}I}{2\pi a}\]
D) \[\frac{{{\mu }_{0}}I}{4\pi a}\sqrt{{{\pi }^{2}}-4}\]
Correct Answer: B
Solution :
\[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I}{a}\times \frac{1}{2}\](due to semicircular part)\[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2I}{a}\](due to parallel parts of currents) These two fields are at right angles to each other. Hence, resultant field \[B=\sqrt{B_{i}^{2}+B_{2}^{2}}=\frac{{{\mu }_{0}}I}{4\pi a}\sqrt{{{\pi }^{2}}+4}\] Hence, the correction option is [b].
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