A 1 m long metal rod AB completes the circui | vedclass

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Question

(D) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B l v$. Given: $B = 0.10 \ T$, $l = 1 \ m$, $v = 1.5 \ m/s$

Vedclass · Accepted Answer

$7.5 	imes 10^{-3}$

Vedclass · Answer

(D) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B l v$. Given: $B = 0.10 \ T$, $l = 1 \ m$, $v = 1.5 \ m/s$, and $R = 2 \ \Omega$. The induced current in the circuit is $I = \frac{\varepsilon}{R} = \frac{B l v}{R}$. The magnetic force acting on the rod is $F_B = I l B = \left( \frac{B l v}{R} ight) l B = \frac{B^2 l^2 v}{R}$. To move the rod with a constant speed, the external force $F_{ext}$ must be equal and opposite to the magnetic force $F_B$. $F_{ext} = F_B = \frac{B^2 l^2 v}{R}$. Substituting the values: $F_{ext} = \frac{(0.1)^2 	imes (1)^2 	imes 1.5}{2} = \frac{0.01 	imes 1 	imes 1.5}{2} = \frac{0.015}{2} = 0.0075 \ N$. $F_{ext} = 7.5 	imes 10^{-3} \ N$.

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