The e.m.f induced at the time the left arm of the frame is at x = 10 cm from the wire is:
JEE Main Online Paper (Held On 19 April 2016)
A) 2 mV
B) 1 mV
C) 0.75 mV
D) 0.5 mV
Correct Answer: B
Solution :
In the given question, Current flowing through the wire, I = 1A Speed of the frame, \[v=10\,m{{s}^{-1}}\] Side of square loop, \[l=10cm\] Distance of square frame from current carrying wires x = 10 cm. We have to find, e.m.f induced e = ? According to Biot-Savart?s law \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{Idl\sin \theta }{{{x}^{2}}}\] \[=\frac{4\pi \times {{10}^{-7}}}{4\pi }\times \frac{1\times {{10}^{-1}}}{{{\left( {{10}^{-1}} \right)}^{2}}}\] \[={{10}^{-6}}\] Induced e.m.f. e = Blv \[={{10}^{-6}}\times {{10}^{-1}}\times 10\] \[=1\mu v\]
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