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 \(................\,10^{-3}\,N\)

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Answer & Solution

Correct Answer

(D) \(18\)

Step-by-step Solution

Detailed explanation

\(F = i \ell B\)

\(=\left(\frac{\varepsilon}{ R }\right) \ell B =\left(\frac{ vB \ell}{ R }\right) \ell B =\frac{ vB ^2 \ell^2}{ R }=\frac{4}{5} \times\left(\frac{15}{100}\right)^2 \times 1^2\)

\(=\frac{4}{5} \times \frac{225}{10^4}\)

\(=\frac{180}{10^4}=0.018\,N\)

\(=18 \times 10^{-3}\,N\)

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