Question

A proton has a speed of 3.0 × 106 m/s in a direction perpendicular to a uniform magnetic field, and the proton moves in a circle of radius 0.20 m. What is the magnitude of the magnetic field?

Answer 1

Answer:

The magnitude of the magnetic field is 0.156 T.

Explanation:

Given that,

Speed of a proton, $v=3\times 10^6\ m/s$

Radius of the circular path, r = 0.2 m

When the proton enters in the circular path, the centripetal force is balanced by the magnetic force such that :

$\dfrac{mv^2}{r}=qvB\sin\theta$

q is the charge on proton

Here, $\theta=90^{\circ}$

$\dfrac{mv^2}{r}=qvB\\\\B=\dfrac{mv}{qr}\\\\B=\dfrac{1.67\times 10^{-27}\times 3\times 10^6}{1.6\times 10^{-19}\times 0.2}\\\\B=0.156\ T$

So, the magnitude of the magnetic field is 0.156 T. Hence, this is the required solution.