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

Answer:

The magnitude of the magnetic field is 10.19 T.

Explanation:

Given that,

Charge on the doubly charged molecule, q = 2e

Magnetic force, $F=6.75\times 10^{-16}\ N$

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

$B=\dfrac{F}{qv}\\\\B=\dfrac{6.75\times 10^{-16}}{2\times 1.6\times 10^{-19}\times 207}\\\\B=10.19\ T$

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