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

Respuesta :

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

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