Question

In the above figure, a circuit is placed perpendicular to a magnetic field $B=0.15 \, T$ . The metal rod, which completes the circuit, is moving with a constant speed of $2ms^{- 1}$ , under the action of a constant force. If the resistance of the total circuit is $3\Omega$ , then find the magnitude of the external force applied:

• A. $3.75\times 10^{- 3}N$
• B. $2.75\times 10^{- 3}N$
• C. $6.57\times 10^{- 4}N$
• D. $4.36\times 10^{- 4}N$

Answer 1

Answer: (A)

The emf induced in the rod causes a current to flow anticlockwise direction in the circuit. Because of this current in the rod, it experiences a force to the left due to the magnetic field. In order to pull the rod to the right with constant sped, this force must be balanced by the puller.
The emf induced in the rod is
$\left|\epsilon \right|=Blv=\left(0.15 \, T\right)\left(0.5 \, m\right)\left(2 \, m \, s^{- 1}\right)=0.15 \, V$
Current induced in the rod is $I=\frac{\left|\epsilon \right|}{R}=\frac{0.15 \, V}{3 \, \Omega}=0.05 \, A$
$\therefore \, F=IlBsin \left(90\right)^{o} = \left(0.05 \, A\right) \left(0.5 \, m\right) \left(0.15 \, T\right) \left(1\right) = 3.75 \times \left(10\right)^{- 3} N$