Question

A magnetic field of \[(4.0\times10^-3 \overrightarrow k)\] T exerts a force of \[(4.0  \overrightarrow i + 3.0 \overrightarrow j ) \times 10^{−10} N\] 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 = `(4xx10xx^-3 hatK)`T

Force exerted by the magnetic field on the charged particle, F = `(4 hati + 3hatj )`× 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 vec i + 100 vec j) `m/s