Question

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 $3\, \Omega$, the force needed to move the rod as indicated with a constant speed of $2 \,m$ will be equal to

• 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 speed, this force must be balanced by the puller The emf induced in the rod is
$|\varepsilon|=B l v=(0.15 T)(0.5 m)\left(2\, m s^{-1}\right)=0.15\, V$
Current induced in the rod is $I=\frac{|\varepsilon|}{R}=\frac{0.15}{3 \Omega}=0.05\, A$
$\therefore F=I l B \sin 90^{\circ}=(0.05\, A)(0.5\, m)(0.15 T)(1)$
$=3.75 \times 10^{-3} N$