A) 0.56 A
B) 5.6 A
C) 0.28 A
D) 2.8 A
Correct Answer: B
Solution :
Magnetic field at the centre of loop. \[B=\frac{{{\mu }_{0}}i}{2r}\] \[\therefore \] \[i=\frac{2Br}{{{\mu }_{0}}}\] \[=\frac{2\times 7\times {{10}^{-5}}\times 5\times {{10}^{-2}}}{4\pi \times {{10}^{-7}}}\] \[=\frac{2\times 7\times {{10}^{-5}}\times 5\times {{10}^{-2}}}{4\times \frac{22}{7}\times {{10}^{-7}}}=5.6A\]
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