Question

How long does it take a proton (mp = 1.67 x 10^-27 kg) moving in an orbit perpendicular to a magnetic field of 2 T to complete one circular orbit?

Answer 1

Answer:
`T=31.42 * 10^(-9) s`

Step-by-step explanation:

Given

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

Magnetic Field(B)=2 T

Required centripetal Force to move in a circle is provided by magnetic field

`mv\omega`=qvB

`\omega =(qB)/(m)`

Where q=charge of proton

B=magnetic Field

v=Velocity of Charge

`\omega`=Angular velocity

Also

`\omega T=2\pi`

`T=(2\pi )/(\omega )`

`T=(2\pi m)/(qB)`

`T=(2* \pi 1.67* 10^(-27))/(1.6* 10^(-19)* 2)`

`T=31.42 * 10^(-9) s`