JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2023
- Medium
- Numerical
A \(20\,cm\) long metallic rod is rotated with \(210\,rpm\) about an axis normal to the rod passing through its one end. The order end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field \(0.2\,T\) parallel to the axis exists everywhere. The emf developed between the centre and the ring is \(.......\,mV\).Take \(\pi=\frac{22}{7}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(88\)
Step-by-step Solution
Detailed explanation
Here \(\omega=210\,rpm\)
\(=210 \times \frac{2 \pi}{60}\,rad / s\)
\(\Rightarrow \omega=7 \pi\,rad / s\)
\(\ell=0.2\,w\)
\(B =0.2\,T\)
\(\text { emf developed across rod is }=\frac{1}{2} B \omega \ell^2\)
\(\frac{1}{2} \times 0.2 \times 7 \pi \times(0.2)^2\)
\(=88\,mV\)
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