Question

A proton, traveling with a velocity of 1.6 × 106 m/s due east, experiences a maximum magnetic force of 8.0 × 10-14 N. The direction of the force is straight down, toward the surface of the earth. What is the magnitude and direction of the magnetic field B, assumed perpendicular to the motion?

Answer 1

Answer:

The magnitude of magnetic field is 0.312 T and direction of the magnetic field is out of the page.

Explanation:

Given:

Velocity of proton $v = 1.6 \times 10^{6}$ $\frac{m}{s}$

Magnetic force $F = 8 \times 10^{-14}$ N

Charge of proton $q = 1.6 \times 10^{-19}$C

The magnetic force experience by proton in magnetic field is given by,

    $F =qvB \sin \theta$

Here magnetic force experienced by proton is maximum so we take $\sin \theta = 1$

    $F = qvB$

    $B = \frac{F}{qv}$

    $B = \frac{8 \times 10^{-14} }{1.6 \times 10^{-19} \times 1.6 \times 10^{6} }$

    $B = 0.312$ T

According to the left hand rule the direction of magnetic field is out of the page.

Therefore, the magnitude of magnetic field is 0.312 T and direction of the magnetic field is out of the page.