Question

9. Alpha particles (charge q = +2e, mass m = 6.6×10−27kg) move at 1.3×10⁶ m/s. What magnetic field strength would be required to bend them into a circular path of radius r = 0.23 m ?

Answer 1

The force on the alpha particle exerted by the magnetic field is given by -

F = Bqv ---------------------------------------------------(i)

Where -

B = Magnetic field strength
q = charge of the particle = 2e = 2 x 1.6 x 10^-19 C
v = velocity of the particle = 1.3 x 10^6 m/s

Now, the force towards the center when moving in a circular path

F = mv^2/r --------------------------------------------(ii)

where -

m = Mass of the particle = 6.6 x 10^-27 kg
v = Velocity
r = Radius = 0.23 m

Equalize equation (i) and (ii) -

Bqv = mv^2/r

=> B x ( 1.6 x 10^-19 x 2 x 1.3 x 10^6) = [ 6.6 x 10^-27 x ( 1.3 x 10^6 )^2 ] / 0.23

=> B =  [ 6.6 x 10^-27 x ( 1.3 x 10^6 ) ] / (0.23 x 2 x 1.6 x 10^-19) = 11.66 x 10^-2 = 0.117 H (Answer)