Question

A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.75 × 10 − 16 N 5.75×10−16 N as it moves at 385 m/s 385 m/s at 63.9 ∘ 63.9∘ to the direction of the field. Find the magnitude of the magnetic field.

Answer 1

Explanation:

The given data is as follows.

           F = $5.75 \times 10^{-16} N$

          q = $1.6 \times 10^{-19} C$

          v = 385 m/s

       $sin (63.9^{o})$ = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

              B = $\frac{F}{qv sin (\theta)}$

                  = $\frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}$

                  = $0.01065 \times 10^{3}$ T

                  = 10.65 T

Thus, we can conclude that magnitude of the magnetic field is 10.65 T.