Question

A proton is sent into a transverse magnetic field of induction 2T with a speed of

3.6x106 m/s. what is the force experienced by it?​

Answer 1

The force experienced by it is 3.204 × 10⁻¹³ Newton

Explanation:

The force experienced by proton moving on the magnetic field is given as:

F = qvB sin θ

Where,

q = Proton = 1.602 × 10⁻¹⁹ C

v = 3.6 × 10⁶ m/s

B = 2 T

On substituting the values, we get,

F = 1.602 × 10⁻¹⁹ × 3.6 × 10⁶ × 2 × sin θ

F = 3.204 × 10⁻¹³ × sin θ

When the proton passes through the magnetic field, the magnetic field, charge and force is mutually perpendicular to each other.

F = 3.204 × 10⁻¹³ × sin 90°

∴ F = 3.204 × 10⁻¹³ N