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  • A square loop of side and resistance <img alt="1 \mathrm{\Omega}" src="https://entrancecorner.codecogs.com/gif.latex?1%20%5Cmathrm%7B%5COmega%7D" /

A square loop of side 20 \mathrm{~cm} and resistance 1 \mathrm{\Omega} is moved towards right with a constant speed v_{0}. The right arm of the loop is in a uniform magnetic field of 5 \mathrm{~T}. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4 \Omega. What should be the value of v_{0} so that a steady current of 2 \mathrm{~mA} flows in the loop ?
Option: 1 \begin{aligned} &10^{-2} \mathrm{~cm} / \mathrm{s} \\ \end{aligned}
Option: 2 1 \mathrm{~cm} / \mathrm{s} \\
Option: 3 1 \mathrm{~m} / \mathrm{s} \\
Option: 4 10^{2} \mathrm{~m} / \mathrm{s}

Answers (1)

best_answer

emf induced in the right arm of the loop is
e= Blv= 5\times 0\cdot 2\times v_{0}
e= 1 v_{0}\rightarrow \left ( 1 \right )
 


I= \frac{e}{4+1}= \frac{v_{0}}{s}= 2mA
\Rightarrow v_{0}= 10\times 10^{-3}\frac{m}{s}
             = 10^{-2}\frac{m}{s}= 1\, \frac{cm}{s}
The correct option is (2)

Posted by

Deependra Verma

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