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

The strength of magnetic field is 0.25 T

Step-by-step explanation:

Time period
`T = 0.262 * 10^(-6)` sec

Mass of proton
`m = 1.67 * 10^(-27)` kg

Charge of proton
`q = 1.6 * 10^(-19)` C

Here proton moves in circular path

`(mv^(2) )/(r) = qvB`

Velocity of proton is given by,

`v = (2\pi r )/(T)`

Put the value of velocity in above equation,

`(m2 \pi r)/(Tr) = qB`

Now magnetic field is given by,

`B = (2\pi m )/(qT)`

`B = ( 6.28 * 1.67 * 10^(-27) )/(1.6 * 10^(-19) * 0.262 * 10^(-6) )`

`B = 0.25` T

Therefore, the strength of magnetic field is 0.25 T