Question

A horizontal wire of mass 'm', length 'l' and resistance 'R' is sliding on the vertical rails on which uniform magnetic field 'B' is directed perpendicular. The terminal speed of the wire as it falls under the force of gravity is ($g =$ acceleration due to gravity)

• A. $\frac{mgl}{BR}$
• B. $\frac{B^{2}l^{2}}{mgR}$
• C. $\frac{mgR}{Bl}$
• D. $\frac{mgR}{B^{2}l^{2}}$

Hint:

Physics Tip: Terminal velocity occurs when the magnetic braking force (proportional to velocity) perfectly balances the constant gravitational force.

Answer 1

The Correct Option is D

Concept: Physics (Electromagnetic Induction) - Motional EMF and Terminal Velocity.

Step 1: Analyze the forces at terminal velocity. The net force on the wire becomes zero when it attains terminal velocity. At this point, the upward magnetic force equals the downward gravitational force: $$ F_{mag} = F_{grav} \text{} $$ $$ iBl = mg \text{} $$

Step 2: Express current in terms of induced EMF. Using Ohm's law, the current $i$ is the induced EMF $e$ divided by resistance $R$: $$ i = \frac{e}{R} \text{} $$ $$ \frac{e}{R}Bl = mg \text{} $$

Step 3: Substitute for motional EMF. The induced motional EMF $e$ for a wire moving with velocity $v$ in a perpendicular magnetic field $B$ is $e = Bvl$. $$ \frac{(Bvl)}{R}Bl = mg \text{} $$ $$ \frac{B^2 v l^2}{R} = mg \text{} $$

Step 4: Solve for terminal velocity $v$. $$ v = \frac{mgR}{B^2 l^2} \text{} $$ $$ \therefore \text{The terminal speed is } \frac{mgR}{B^{2}l^{2}}. \text{} $$