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Question

(A) The motional electromotive force $(EMF)$ induced in the rod is $\varepsilon = B l v$. The total resistance of the circuit at distance $x$ is $R_{t

Vedclass · Accepted Answer

$i l B + \frac{2m \lambda i^2}{B^2 l^2}(R + 2 \lambda x)$

Vedclass · Answer

(A) The motional electromotive force $(EMF)$ induced in the rod is $\varepsilon = B l v$. The total resistance of the circuit at distance $x$ is $R_{total} = R + 2 \lambda x$. Since the current $i$ is constant,we have $i = \frac{B l v}{R + 2 \lambda x}$,which implies $v = \frac{i(R + 2 \lambda x)}{B l}$. Differentiating $v$ with respect to $x$,we get $\frac{dv}{dx} = \frac{i}{B l} \cdot \frac{d}{dx}(R + 2 \lambda x) = \frac{2 \lambda i}{B l}$. The acceleration $a$ of the rod is $a = v \frac{dv}{dx} = \left[ \frac{i(R + 2 \lambda x)}{B l} \right] \left( \frac{2 \lambda i}{B l} \right) = \frac{2 \lambda i^2 (R + 2 \lambda x)}{B^2 l^2}$. Applying Newton's second law to the rod,the net force is $F - F_{mag} = ma$,where $F_{mag} = i l B$. Thus,$F = i l B + ma = i l B + \frac{2m \lambda i^2}{B^2 l^2}(R + 2 \lambda x)$.

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