First slide

Energy consideration quantitative study.

Question

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

Moderate

A

$3.75 \times 10^{- 3} N$
B

$3.75 \times 10^{- 2} N$
C

$3.75 \times 10^{2} N$
D

$3.75 \times 10^{- 4} N$

Solution

Induced current in the circuit, $i = \frac{Bvl}{R}$
Magnetic force acting on the wire,
$F_{m} = Bil = B (\frac{Bvl}{R}) l$
$\Rightarrow F_{m} = \frac{B^{2} {vl}^{2}}{R}$
External force needed to move the rod with constant velocity is
$(F_{m}) = \frac{B^{2} {vl}^{2}}{R} = \frac{(0.15)^{2} \times (2) \times (0.5)^{2}}{3}$
$= 3.75 \times 10^{- 3} N$

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