Magnetic field due to currentMedium
Question
A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T . If the resistance of the total circuit is $2\Omega$ then the force needed to move the rod towards right with constant speed (v) of $1.5\text{ }m/s$ is $\_\_\_\_$ N.
Options
A.$7.5 \times 10^{- 2}$
B.$5.7 \times 10^{- 3}$
C.$5.7 \times 10^{- 2}$
D.$7.5 \times 10^{- 3}$
Solution
To maintain constant speed
$$\begin{matrix} & F_{ext} = F_{B} \\ & \ \Rightarrow {\text{ }F}_{ext} = ilB \\ & \ = \left( \frac{vBl}{R} \right)l\text{ }B \\ & \ = \frac{B^{2}l^{2}v}{R} \\ & \ = \frac{(0.1)^{2} \times (1)^{2} \times 1.5}{2} \\ & \ = 7.5 \times 10^{- 3}\text{ }N \end{matrix}$$
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