Neet Solved Paper 2015 Question 31
Question: 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
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Options:
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}) $
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Answer:
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}) $
