Question

A positive charge of 0.049 C moves horizontally to the right at a speed of 272.4 m/s and enters a magnetic field directed vertically downward. If it experiences a force of 22.394 N, what is the magnetic field strength ?

Answer 1

The magnitude of the force F experienced by a charge q that moves in a direction perpendicular to a magnetic field B with a speed v is given by:

$F=qvB$

Isolate B from the equation:

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

Replace F=22.394N, q=0.049C and v=272.4m/s to find the strength of the magnetic field:

$B=\frac{22.394N}{(0.049C)(272.4\frac{m}{s})}=1.6778...T\approx1.7T$

Therefore, the magnetic field strength is approximately 1.7T.