Question

A proton with a speed of 3.2 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.23 m. As the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field so the proton just misses colliding with the opposite plate

Answer 1

The magnitude of the magnetic is 0.145 T

Step-by-step explanation:

Given :

Speed of proton
`v = 3.2 * 10^(6)(m)/(s)`

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

The force on the proton in magnetic field is given by,

`F = q (v * B)`

`F = qvB \sin \theta`

But
`\sin 90 = 1` (∵ Force is perpendicular to the velocity so
`\theta = 90`)

`F = qvB`

When particle enter in magnetic field at the angle of 90° so particle moves in circle

So force is given by,

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

Where
`r =` radius but in our case 0.23 m,
`q = 1.6 * 10^(-19)` C

By comparing above two equation,

`B = (mv)/(qr)`

`B = (1.67 * 10^(-27) * 3.2 * 10^(6) )/(1.6 * 10^(-19) * 0.23 )`

`B = 0.145` T