# A proton moves in a circular path perpendicular to a constant magnetic field so that it takes 0.262 × 10−6 s to complete the revolution. Det

Question

A proton moves in a circular path perpendicular to a constant magnetic field so that it takes 0.262 × 10−6 s to complete the revolution. Determine the strength of the constant magnetic field. The angular speed is given in radians per unit time

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2 years 2021-07-27T15:15:07+00:00 1 Answers 29 views 0

The strength of magnetic field is 0.25 T

Explanation:

Time period $$T = 0.262 \times 10^{-6}$$ sec

Mass of proton $$m = 1.67 \times 10^{-27}$$ kg

Charge of proton $$q = 1.6 \times 10^{-19}$$ C

Here proton moves in circular path

$$\frac{mv^{2} }{r} = qvB$$

Velocity of proton is given by,

$$v = \frac{2\pi r }{T}$$

Put the value of velocity in above equation,

$$\frac{m2 \pi r}{Tr} = qB$$

Now magnetic field is given by,

$$B = \frac{2\pi m }{qT}$$

$$B = \frac{ 6.28 \times 1.67 \times 10^{-27} }{1.6 \times 10^{-19} \times 0.262 \times 10^{-6} }$$

$$B = 0.25$$ T

Therefore, the strength of magnetic field is 0.25 T