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 m/s is $\_\_\_\_$ N.

Question

Accepted Answer

D. 1. When a rod of length $l$ moves with velocity $v$ perpendicular to a field $B$, it develops a **motional emf**.
$$
\varepsilon = Bvl
$$
2. Substitute $B=0.10$ T, $v=1.5$ m/s, $l=1$ m.
$$
\varepsilon = (0.10)	imes(1.5)	imes(1) = 0.15	ext{ V}
$$
3. Use **Ohm's Law** to find the induced current, given total resistance $R=2\Omega$.
$$
I = \frac{\varepsilon}{R} = \frac{0.15}{2} = 0.075	ext{ A}
$$
4. A current-carrying rod in a magnetic field experiences a **force** opposing its motion (by Lenz's Law), given by $F_m=IlB$ since the rod is perpendicular to the field.
$$
F_m = (0.075)	imes(1)	imes(0.10) = 0.0075	ext{ N} = 7.5	imes10^{-3}	ext{ N}
$$
5. Since the rod moves at **constant speed**, the net force is zero, so the applied external force must exactly balance this magnetic braking force.
$$
F_{ext} = F_m = 7.5	imes10^{-3}	ext{ N}
$$
Hence, the force needed is $7.5	imes10^{-3}$ N, so the answer is **option D**.