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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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Physics
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2021-07-27T15:15:07+00:00
2021-07-27T15:15:07+00:00 1 Answers
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Answer:
The strength of magnetic field is 0.25 T
Explanation:
Time period [tex]T = 0.262 \times 10^{-6}[/tex] sec
Mass of proton [tex]m = 1.67 \times 10^{-27}[/tex] kg
Charge of proton [tex]q = 1.6 \times 10^{-19}[/tex] C
Here proton moves in circular path
[tex]\frac{mv^{2} }{r} = qvB[/tex]
Velocity of proton is given by,
[tex]v = \frac{2\pi r }{T}[/tex]
Put the value of velocity in above equation,
[tex]\frac{m2 \pi r}{Tr} = qB[/tex]
Now magnetic field is given by,
[tex]B = \frac{2\pi m }{qT}[/tex]
[tex]B = \frac{ 6.28 \times 1.67 \times 10^{-27} }{1.6 \times 10^{-19} \times 0.262 \times 10^{-6} }[/tex]
[tex]B = 0.25[/tex] T
Therefore, the strength of magnetic field is 0.25 T