Question

A proton moves perpendicular to a uniform magnetic field B at a speed of 1.30 x 10^7 m/s and experiences an acceleration of 2.40 x 10^13 m/s^2 in the positive x direction when its velocity is in the positive z direction.  Determine the magnitude and direction of the field?

Answer 1

1) The magnetic force exerted on the proton, since it is moving perpendicular to the magnetic field, is
F=qvB
where q is the proton charge, v its velocity and B the intensity of the magnetic field.
Since the proton is moving by circular motion, this force is equal to the centripetal force:
F=qvB=ma_c 
where m is the proton mass and ac is the centripetal acceleration.

Substituting the data of the problem and re-arranging the formula, we find:\(B=\frac{m a_{c}}{q v}=\frac{\left(1.67 \cdot 10^{-27} k g\right)\left(2.40 \cdot 10^{13} \mathrm{~m} / \mathrm{s}^{2}\right)}{\left(1.6 \cdot 10^{-19} \mathrm{C}\right)\left(1.30 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\right)}=0.01924 T\) so, the magnitude of the magnetic field is 0.019 T.

2) Direction:
the direction of the magnetic field can be found by using the right-hand rule. 
Let's take:
- the index finger as the direction of the velocity (positive z-direction)
- the middle finger as the direction of the magnetic field
- the thumb as the direction of the force (which has same direction as the acceleration) (positive x-direction)
we can see that by using the right hand, the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction.

The magnitude of magnetic field is 0.019 T and diection is toword negative y axis.