An electron moves to the left with a velocity of 9.6 x 10^5 m/s and has a magnetic force of 2.6 x 10^-15 n exerted on it. what is the magnitude of the magnetic field? assume the charge has a value of 1.60 x 10^-19 c.

The magnitude of the magnetic field strength is  16.927 mT

Explanation:

Given;

velocity of the electron, v = 9.6 x 10⁵ m/s

magnitude of the force exerted on the electron, F =  2.6 x 10⁻¹⁵ N

charge of electron, q = 1.60 x 10⁻¹⁹ C

The force exerted on the electron in magnetic field is given by charge times the vector product of velocity and magnetic field strength.

F = q(v x B)

where;

B is the magnetic field strength

Substitute the given values of force, charge and velocity and calculate the magnetic field strength:

Therefore, the magnitude of the magnetic field strength is 16.927 mT

unlike poles attract

small = small current large mag large current