Question

What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.8 t ?

Answer 1

The magnetic force acting on the proton is 
$F=qvB \sin \theta$
where
q is the proton charge
v is its speed
B is the intensity of the magnetic field
$\theta$ is the angle between the direction of v and B; since the proton is moving perpendicular to the magnetic field, $\theta=90^{\circ}$ and $\sin \theta=1$, so the force becomes
$F=qvB$

this force provides the centripetal force that keeps the proton in circular motion:
$m \frac{v^2}{r} = q v B$
where the term on the left is the centripetal force, with
m being the mass of the proton
r the radius of its orbit

Re-arranging the previous equation, we can find the radius of the proton's orbit:
$r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(6.5 m/s)}{(1.6 \cdot 10^{-19} C)(1.8 T)}=3.77 \cdot 10^{-8}m$

And now we can calculate the centripetal acceleration of the proton, which is given by
$a_c = \frac{v^2}{r}= \frac{(6.5 m/s)^2}{3.77\cdot 10^{-8}m}=1.12 \cdot 10^9 m/s^2$