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?

The magnitude of the magnetic field is 0.156 T.

Explanation:

Given that,

Speed of a proton,

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 :

q is the charge on proton

Here,

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

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