A) A
B) 0.2 A
C) 0.3 A
D) 0.4 A
Correct Answer: A
Solution :
[a] : Here, total length \[l=40\text{ }cm=40\times {{10}^{-}}^{2}m\], Resistivity \[=1.7\times {{10}^{-8}}\Omega m\] The area A of the loop \[=\left( \frac{40cm}{4} \right)\left( \frac{40cm}{4} \right)=0.01{{m}^{2}}\] If the magnetic field at an instant is B, the flux through the frame at that instant will be \[\phi =BA\]. As the area remains constant, the magnitude of the emf induced will be\[\varepsilon =\frac{d\phi }{dt}=A\frac{dB}{dt}\] \[=(0.01\text{ }{{m}^{2}})(0.02\,T\,{{\text{s}}^{-1}})=2\times {{10}^{-4}}V\] Resistance of the loop, \[R=\frac{(1.7\times {{10}^{-8}}\Omega m)(40\times {{10}^{-2}}m)}{3.14\times 1\times {{10}^{-6}}{{m}^{2}}}=2.16\times {{10}^{-3}}\Omega \] Hence, the current induced in the loop will be \[I=\frac{2\times {{10}^{-4}}V}{2.16\times {{10}^{-3}}\Omega }=9.3\times {{10}^{-2}}A\approx 0.A\]
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