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.
Answer :
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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