Question

I got the first part, but am stuck on the 2nd! Thanks in advance!

A proton has a magnetic field due to its spin on its axis. The field is similar to that created by a circular current loop 0.646 x10–15 m in radius with a current of 1.05 x104 A. Find the maximum torque on a proton in a 2.53 T field. 3.48 X10-26 Nm In the previous problem, we calculated the torque on a proton as it spins about its internal axis of rotation. (Hint: This motion is analogous to Earth spinning about its poles every 24 hours.) If the average radius of a proton is 0.85 x10–15 m, what is the angular acceleration caused by the proton being in the magnetic field? The mass of a fa proton is 1.67 x10-27 kg. X10+31 m/s/s Note that ~2-3 Tesla is a large B-field, especially for a tiny proton!

Answer 1

2) So we know that in rotational kinematics

$\tau =I\alpha$

here I is moment of inertia of proton which we can treat as solid sphere and

alpha is angular accelaration of proton

and Tau is torque on proton

so,

$\tau =\frac{2}{5}mR^2\alpha$

so i am assuming torque calculated by you was correct then

$\alpha =\frac{5\tau}{2mR^2}$

R is radius of proton

m is mass of proton

so plug in all values we get

$\alpha =7.21 \times 10^{31}rad/s^2$