A) \[4\times {{10}^{-6}}T\], vertical up
B) \[5\times {{10}^{-6}}T\], vertical down
C) \[5\times {{10}^{-6}}T\], vertical up
D) \[4\times {{10}^{-6}}T\], vertical down
Correct Answer: C
Solution :
(c) \[5\times {{10}^{-6}}T\], vertical up From Ampere circuital law \[\oint{\overrightarrow{B}\cdot \overrightarrow{dl}={{\mu }_{0}}{{l}_{enc}}}\] \[B\times 2\pi R={{\mu }_{0}}{{l}_{enc}}\] \[B=\frac{{{\mu }_{0}}{{l}_{enc}}}{2\pi R}=2\times {{10}^{-7}}\times \frac{75}{3}=5\times {{10}^{-6}}T\] The direction of field at the given point will be vertical up determined by the screw rule or right-hand rule.
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