Q. 3

# A magnetic field of (4.0 x 10-3 vector k)T exerts a force of (4.0 vector i + 3.0 vector j) x 10-10N on a particle having a charge of 1.0 x 10-9C and going in the X-Y plane. Find the velocity of the particle.

Given -

Magnetic field, B = 4.0 × 10–3
T

Electric charge on the particle, q = 1 × 10−9 C

Also given that the charge is going in the X-Y plane,

Therefore, the x-component of force, Fx = 4 × 10−10 N

and the y-component of force, Fy = 3 × 10−10 N

We know, Lorentz force F is given by -

where,

q = charge

v = velocity of the charge

B=magnetic field

and

θ= angle between V and B

On putting the respective values, we get

Let’s take motion along x-axis-

From (1)

Similarly, motion along y-axis

from (1)

Hence. velocity of the particle, = m/s

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