A) 1.44 A
B) 0.793 A
C) 2.88 A
D) 3.4 A
Correct Answer: B
Solution :
Suppose the magnetic fields due to big and small coils are By and B^ respectively. \[\therefore \] \[{{B}_{1}}=\frac{{{\mu }_{0}}NI}{2{{a}_{1}}}\] Or \[{{B}_{1}}=\frac{4\pi \times {{10}^{-7}}\times 100\times 5}{2\times 0.07}\] Or \[{{B}_{1}}=4.49\,mT\] \[{{B}_{2}}=\frac{{{\mu }_{0}}NI}{2{{a}_{2}}}=\frac{4\pi \times {{10}^{-7}}\times 100\times {{I}_{2}}}{2\times 0.02}\] Again, \[B={{B}_{1}}-{{B}_{2}}=2\] \[{{B}_{2}}={{B}_{1}}-B\] \[=4.49-2=2.49\text{ }mT\] Now, \[{{I}_{2}}=\frac{2.49\times {{10}^{-3}}\times 2\times 0.02}{4\pi \times {{10}^{-7}}\times 100}\] \[=0.793A\]
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