| A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plate of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of \[10m{{s}^{-1}}\] (see figure). |
 |
| The e.m.f induced at the time the left arm of the frame is at x = 10 cm from the wire is: [JEE ONLINE 19-04-2014] |
A) 2 mV
B) 1 mV
C) 0.75 mV
D) 0.5 mV
Correct Answer: B
Solution :
| [b] 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-Savarts 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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