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

Answer:

The magnitude of the magnetic is 0.145 T

Explanation:

Given :

Speed of proton $v = 3.2 \times 10^{6}\frac{m}{s}$

Mass of proton $m = 1.67 \times 10^{-27}$ Kg

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

  $F = q (v \times B)$

  $F = qvB \sin \theta$

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,

  $F = \frac{mv^{2} }{r}$

Where $r =$ radius but in our case 0.23 m, $q = 1.6 \times 10^{-19}$ C

By comparing above two equation,

  $B = \frac{mv}{qr}$

  $B = \frac{1.67 \times 10^{-27} \times 3.2 \times 10^{6} }{1.6 \times 10^{-19} \times 0.23 }$

  $B = 0.145$ T