As shown in figure, a metal rod completes the circuit. The circuit area is perpendicular to a magnetic field with $B = 0.15 T$ . If the resistance of the total circuit is $3 Ω$ , how large a force is needed to move the rod as indicated with a constant speed of $2 m / s$ ?

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Solution

$e = B v l$
$i = \frac{e}{R} = \frac{B v l}{R}$
$F = i l B = \frac{B^{2} l^{2} v}{R}$
$= \frac{{(1.5)}^{2} {(0.5)}^{2} (2)}{3}$
$= 0.00375 N$

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Q. As shown in the figure, a metal rod makes contact with a partial circuit and completes the circuit. The Circuit area is perpendicular to a magnetic field with B = 0.15 T. If the resistance of the total circuit is

3 Ω

, the force needed to move the rod as indicated with a constant speed of 2 m

s^{- 1}

will be equal to

As shown in figure, a metal rod makes contact with a partial circuit and completes the circuit. The circuit area is perpendicular to a magnetic field with B = 0.15 T. If the resistance of the total circuit is $3 Ω$ , the force needed to move the rod as indicated with a constant speed of $2 m s^{- 1}$ will be equal to