As shown in the figure,a metal rod makes cont | vedclass

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Question

(A) The induced electromotive force $(EMF)$ in the rod is given by $\varepsilon = Bvl$,where $B = 0.15 \, T$,$v = 2 \, m/s$,and $l = 50 \, cm = 0.5 \,

Vedclass · Accepted Answer

$3.75 	imes 10^{-3} \, N$

Vedclass · Answer

(A) The induced electromotive force $(EMF)$ in the rod is given by $\varepsilon = Bvl$,where $B = 0.15 \, T$,$v = 2 \, m/s$,and $l = 50 \, cm = 0.5 \, m$.The induced current in the circuit is $i = \frac{\varepsilon}{R} = \frac{Bvl}{R}$,where $R = 3 \, \Omega$.The magnetic force acting on the rod is $F_m = Bil = B \left( \frac{Bvl}{R} ight) l = \frac{B^2 v l^2}{R}$.To move the rod at a constant speed,the external force $F_{ext}$ must be equal in magnitude to the magnetic force $F_m$.$F_{ext} = \frac{(0.15)^2 	imes 2 	imes (0.5)^2}{3} = \frac{0.0225 	imes 2 	imes 0.25}{3} = \frac{0.01125}{3} = 0.00375 \, N = 3.75 	imes 10^{-3} \, N$.

As shown in the figure,a metal rod makes contact and completes the circuit. The circuit is perpendicular to the magnetic field with $B = 0.15 \, T$. If the resistance is $3 \, \Omega$,the force needed to move the rod as indicated with a constant speed of $2 \, m/s$ is:

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