Question

A charged particle moving through a magnetic field at right angles to the field with a speed of 35.1 m/s experiences a magnetic force of 7.56x10-4 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 8.7 m/s at an angle of 44° relative to the magnetic field. Express your answer in microNewtons.

Answer 1

Magnetic force on charged particle in magnetic field is given by,

F = q*v*B*sin

here in first case,

F = magnetic force = 7.56*10^-4 N

q = charge on particle

B = magnetic field strength = ??

v = speed of particle = 35.1 m/s

= 90 deg (given, moving at right angle)

So,

B = F/(q*v*sin)

B = (7.56*10^-4)/(q*35.1*sin(90 deg))

Now, in second case:

v' = new speed of particle = 8.7 m/s

' = 44 deg (given)

q' = q (As, both particl are identical)

So, new magnetic force in same magnetic field will be:

F' = q'*v'*B*sin'

F' = q*8.7*[(7.56*10^-4)/(q*35.1*sin(90 deg))]*sin(44 deg)

F' = 8.7*sin(44 deg)*(7.56*10^-4)/(35.1*sin(90 deg))

F' = 1.30*10^-4 N = 130*10^-6 N

F' = 130 N

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