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JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A \(1\,m\) long metal rod \(XY\) completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density \(0.15\,T\). If the resistance of the circuit is \(5\,\Omega\), the force needed to move the rod in direction, as indicated, with a constant speed of \(4\,m / s\) will be \(................\,10^{-3}\,N\)
- A \(9\)
- B \(45\)
- C \(16\)
- D \(18\)
Answer & Solution
Correct Answer
(D) \(18\)
Step-by-step Solution
Detailed explanation
\(F = i \ell B\) \(=\left(\frac{\varepsilon}{ R }\right) \ell B =\left(\frac{ vB \ell}{ R }\right) \ell B =\frac{ vB ^2 \ell^2}{ R }=\frac{4}{5} \times\left(\frac{15}{100}\right)^2 \times 1^2\) \(=\frac{4}{5} \times \frac{225}{10^4}\) \(=\frac{180}{10^4}=0.018\,N\)…
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