Question

A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 6.75 × 10 − 16 N 6.75×10−16 N as it moves at 207 m/s 207 m/s at 74.9 ∘ 74.9∘ to the direction of the field. Find the magnitude of the magnetic field.

Answer 1

The magnitude of the magnetic field is 10.19 T.

Step-by-step explanation:

Given that,

Charge on the doubly charged molecule, q = 2e

Magnetic force,
`F=6.75* 10^(-16)\ N`

Speed of molecule, v = 207 m/s

The angle to the direction of the field is 74.9 degrees.

We need to find the magnitude of magnetic field. The formula of magnetic force is given by :

`F=qvB`

B is magnetic field

`B=(F)/(qv)\\\\B=(6.75* 10^(-16))/(2* 1.6* 10^(-19)* 207)\\\\B=10.19\ T`

So, the magnitude of the magnetic field is 10.19 T.