Question

A magnetic field of (4.0 × 10⁻³ $\overrightarrow{k}$) T exerts a force of (4.0 $\overrightarrow{i}$ + 3.0 $\overrightarrow{j}$) × 10⁻¹⁰ N on a particle with a charge of 1.0 × 10⁻⁹ C and going in the x−y plane. Find the velocity of the particle.

Answer 1

Given:
Magnetic field, B = (4 × 10⁻³ $\hat{k}$)T
Force exerted by the magnetic field on the charged particle, F = (4 $\hat{i}$ + 3 $\hat{j}$) × 10⁻¹⁰ N
Charge of the particle, q = 1 × 10⁻⁹ C
As per the question, the charge is going in the X-Y plane.
So, the x-component of force, F_x = 4 × 10⁻¹⁰ N
and the y-component of force, F_y = 3 × 10⁻¹⁰ N
Considering the motion along x-axis:
F_x = qv_y×B
On putting the respective values, we get:
v_y = 100 m/s
Motion along y-axis:
F_y = qv_x×B
⇒ v_x = 75 m/s
Thus, total velocity = (−75 $\overrightarrow{i}$ + 100 $\overrightarrow{j}$) m/s