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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