Question

A deuteron particle (the nucleus of an isotope of hydrogen consisting of one proton and one neutron and having a mass of 3.34×10−27 kg ) moving horizontally enters a uniform, vertical, 0.600 T magnetic field and follows a circular arc of radius 56.5 cm .

Part A

How fast was this deuteron moving just before it entered the magnetic field ?

Part B

How fast was this deuteron moving just after it came out of the field?

Part C

What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuteron?

Answer 1

Part A -

Here, the force exerted on the deuteron F = q*V*B

And the centripetal force F = M*V^2/r

Equalize the both -

q*V*B=m*V^2/r
q*B=m*V/r

=> V = q*r*B/m = (1.6x10^-19 x 0.565 x 0.60 ) / (3.34 x 10^-27) = 1.62 x 10^7 m/s

Part B -  

The deuteron would be going the same speed (1.62 x 10^7 m/s) as it leaves the field, because the lorentz force always acts perpendicular to the velocity: it can only change direction, not speed.

Part C -

In the case of proton, all the variables will remain the same except -

mass of proton, m = 1.67 x 10^-27 kg

So, r = m*V/(q*B) = (1.67 x 10^-27 x 1.62 x 10^7) / (1.6 x 10^-19 x 0.60) = 2.82 x 10^-1 m = 0.282 m = 28.2 cm