A) \[{{10}^{-17}}\,{{\mu }_{0}}\]
B) \[{{10}^{-14}}\,{{\mu }_{0}}\]
C) \[{{10}^{-7}}\,{{\mu }_{0}}\]
D) \[{{10}^{-5}}\,{{\mu }_{0}}\]
Correct Answer: A
Solution :
For a coil of radius a, carrying current i, magnetic field at the centre is given -by \[B=\frac{{{\mu }_{0}}i}{2\,a}N{{A}^{-1}}{{m}^{-1}}\] Also, current \[i=\frac{charge\text{ }\left( q \right)}{time\,(t)}\]= change \[\times \]frequency Given, \[q=100\,e=100\times 1.6\times {{10}^{-19}}C\] \[=1.6\times {{10}^{-17}}C\] \[f=1\,rot/s\]. \[\therefore \] \[i=1.6\times {{10}^{-17}}A\] Hence, \[B={{\mu }_{0}}\times \frac{1.6\times {{10}^{-17}}}{2\times 0.8}={{10}^{-17}}{{\mu }_{0}}\]
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