A) 1.06 A
B) 2.44 A
C) 3.44 A
D) 4.44 A
Correct Answer: A
Solution :
(a) 1.06 A Here, \[N=90\] \[R=15\,cm=15\times {{10}^{-2}}m,\,B=4\times {{10}^{-4}}T\] \[\because B=\frac{{{\mu }_{0}}Nl}{2R}\] \[\therefore \,\,\,\,\,\,\,\,l=\frac{2RB}{N{{\mu }_{0}}}=\frac{2\times 15\times {{10}^{-2}}\times 4\times {{10}^{-4}}}{4\pi \times {{10}^{-7}}\times 90}=1.06A\]
You need to login to perform this action.
You will be redirected in
3 sec
