Question

a doubly ionized molecule, moving into a nagnetic field, experiences a magnetic force of 6.01×10^-16 N as it moves at 251m/s at 67.5° to the direction of the field. find the magnitude of the magnetic field in T.

Answer 1

Given, charge on the molecule, q = - 2 * 1.6 * 10^-19 C

Velocity of the molecule, v = 251 m/s

Magnetic force on the molecule, F = 6.01 * 10^-16 N

Angle between velocity and direction of magnetic field, = 67.5^°​​​​​​

Magnetic force, F on a moving particle with velocity, v in a magnetic field B is given by :

F = q*v*B*sin​​​​​​

B = F/(q*v*sin​​​​​​)

B = 8.11 T