Question

A ball of mass 100 gm and carrying charge 1 C is released from a point near the surface of earth, in a uniform horizontal magnetic field of 2T. Find the maximum speed attained by the ball in m/s. (Take g = 10 m/s², assume ball attains maximum speed before hitting ground)

• A. 1.5 m/s
• B. 1 m/s
• C. 2 m/s
• D. 0.5 m/s

Answer 1

Answer: (D)

0.5 m/s

Answer 2

The magnetic force on the ball is

$$F_{\text{mag}} = qvB,$$

while the gravitational force is

$$F_g = mg.$$

Since the magnetic force acts perpendicular to the velocity, it does no work. However, if we suppose that the ball attains a state where the magnetic force “matches” the gravitational force in magnitude (i.e. the magnetic deflection becomes significant) before it hits the ground, we can set

$$qvB = mg.$$

Solving for $v$:

$$v = \frac{mg}{qB} = \frac{0.1\,\mathrm{kg} \times 10\,\mathrm{m/s^2}}{1\,\mathrm{C} \times 2\,\mathrm{T}} = \frac{1}{2}\,\mathrm{m/s} = 0.5\,\mathrm{m/s}.$$