Electromagnetic Induction Part 2

Concept:When a conductor moves perpendicular to a uniform magnetic field, a motional EMF is induced. The resulting current experiences a magnetic force that opposes the motion. To maintain constant velocity, an equal external force must be applied.Explanation:Given: $l = 1$ m, $B = 0.10$ T, $v = 1.5$ m/s, $R = 2$ $\Omega$.Motional EMF: $\varepsilon = B l v = (0.10)(1)(1.5) = 0.15$ V.Induced current (Ohm’s law): $I = \varepsilon / R = 0.15 / 2 = 0.075$ A.Magnetic force on the current-carrying rod: $F_m = I l B \sin\theta$. Since the rod is perpendicular to the field ($\theta = 90^\circ$), $\sin\theta = 1$.Thus $F_m = I l B = (0.075)(1)(0.10) = 0.0075$ N.By Lenz’s law, this force acts opposite to the motion (to the left).For constant speed, net force must be zero. Therefore, external force required: $F_{\text{ext}} = F_m = 7.5 \times 10^{-3}$ N.Answer:Option B: $\mathbf{7.5 \times 10^{-3}}$ N.