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Anuta_ua [19.1K]
3 years ago
12

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?
Biology
1 answer:
jeyben [28]3 years ago
3 0

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.

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