JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2019
- Hard
- Numerical
The figure shows a square loop \(L\) of side \(5\, cm\) which is connected to a network of resistances. The whole set up is moving towards right with a constant speed of \(1\, cms^{-1}\). At some instant, a part of \(L\) is in a uniform magnetic field of \(1\, T\), perpendicular to the plane of the loop. If the resistance of \(L\) is \(1.7\,\Omega \), the current in the loop at that instant will be close to.....\(\mu A\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(170\)
Step-by-step Solution
Detailed explanation
since it is a balanced wheatstone bridge, its equivalent resistance \(=\frac{4}{3}\, \Omega\)
\(\varepsilon=\mathrm{B} \ell \mathrm{v}=5 \times 10^{-4}\, \mathrm{V}\)
So total resistance
\(\mathrm{R}=\frac{4}{3}+1.7 \approx 3\, \Omega\)
\(\therefore \quad \mathrm{i}=\frac{\varepsilon}{\mathrm{R}} \approx 166\, \mu \mathrm{A} \approx 170\, \mu \mathrm{A}\)
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