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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.
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Solution
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, Fx = 4 × 10−10 N
and the y-component of force, Fy = 3 × 10−10 N
Considering the motion along x-axis:-
Fx = qvy×B
On putting the respective values, we get:-
vy = 100 m/s
Motion along y-axis:-
Fy = qvx×B
⇒ vx = 75 m/s
Thus, total velocity =`( -75 vec i + 100 vec j) `m/s
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