JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2022
- Medium
- Numerical
As shown in the figure, a metallic rod of linear density \(0.45\,kg\,m ^{-1}\) is lying horizontally on a smooth incline plane which makes an angle of \(45^{\circ}\) with the horizontal. The minimum current flowing in the rod required to keep it stationary, when \(0.15\,T\) magnetic field is acting on it in the vertical upward direction, will be \(....A\) \(\left\{\right.\) Use \(\left.g=10 m / s ^{2}\right\}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(30\)
Step-by-step Solution
Detailed explanation
\(mg \sin 45^{\circ}= ILB \cos 45^{\circ}\)
\(\therefore I =\left(\frac{ m }{ L }\right) \frac{ g }{ B }\)
\(=\frac{(0.45)(10)}{0.15}=30\,A\)
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