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

Answer 1

Answer:

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