Question

A proton, traveling with a velocity of 4.5 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.0 × 10−14 N and a direction of due south. What are the magnitude and direction of the magnetic field causing the force?

Answer 1

Answer:

Magnetic field, B = 0.11 i (in k direction)

Explanation:

It is given that,

Velocity of the proton, $v=4.5\times 10^6\ m/s$

Magnetic force, $F=8\times 10^{-14}\ N$ (due south)

The magnetic force acting on the electron is given by :

$F=qvB$

(-j) = (i) × (B)

$B=\dfrac{F}{qv}$

q is the charge on proton

$B=\dfrac{8\times 10^{-14}\ N}{1.6\times 10^{-19}\ C\times 4.5\times 10^6\ m/s}$

B = 0.11

So, the magnitude of 0.11 T is acting on the proton in and is acting directed upward above the plane. Hence, this is the required solution.