Question

A proton moving in a circular path perpendicular to a constant magnetic field takes 2.6 µs (2.6 × 10−6 s) to complete one revolution. The proton has a charge of 1.60218 × 10−19 C and a mass of 1.67262 × 10−27 kg. Determine the magnitude of the field. Answer in units of T.

Answer 1

Answer:

The magnitude of the field is 0.025 Tesla.  

Explanation:

The magnetic field force over a charge particle is defined as:

$F_{B} = qvB sen \theta$  (1)

             

Where q is the charge, v is the velocity, B is the magnetic field and $\theta$ is the angle between the displacement of the particle and the magnetic field.

For this particular case, $\theta$ is equal to $90^\circ$, since the proton is moving in a circular path perpendicular to a constant magnetic field.

Therefore, from equation 1 it is gotten:

$F_{B} = qvB$  (2)

Then, replacing Newton's second law in equation 2 it is gotten:

                     

$ma = qvB$

However, a is the centripetal acceleration since the proton describes a circular motion:

$m\frac{v^{2}}{r} = qvB$ (3)

Notice that the magnetic field, B, can be isolated from equation 3

$B = \frac{mv^{2}}{qvr}$

$B = \frac{mv}{qr}$   (4)

But v can be replaced with the orbital speed:

$v = \frac{2 \pi r}{T}$  (5)

Where T is the period.

Therefore, equation 5 can be replaced in equation 4

$B = \frac{m2 \pi r}{Tqr}$

$B = \frac{2 \pi m}{Tq}$  (6)

Finally, equation 6 can be used with the values given:

$B = \frac{2 \pi (1.67262x10^{-27}Kg)}{(2.6x10^{-6} s)(1.60218x10^{-19} C)}$         

But $1C = 1 A.s$ and $1T = Kg^{2}.s^{-2}.A^{-1}$    

$B = \frac{2 \pi (1.67262x10^{-27}Kg)}{(2.6x10^{-6} s)(1.60218x10^{-19} A.s)}$        

$B = 0.025 T$  

Hence, the magnitude of the field is 0.025 Tesla.