Question

A proton enters into a magnetic field of induction 1.732 T, with a velocity of 10⁷ m/s at an angle 60° to the field. The force acting on the proton is e = 1.6 × 10^-19 C, sin 60° = cos 30° = `sqrt3/2`

Options

• 1.6 × 10^-12 N

• 3.2 × 10^-12 N

• 4.0 × 10^-12 N

• 2.4 × 10^-12 N

Answer 1

2.4 × 10^-12 N

Explanation:

F = q(B × v) = qBv sin 60° = 1.732 × 1.6 × 10^-19 × 10⁷ × `sqrt3/2`

= 2.4 × 10^-12 N