Answer:
[tex]7.04\times 10^{-16} N[/tex]
Explanation:
We are given that
Current,I=66 A
Speed,v=[tex]1.0\times 10^7 m/s[/tex]
d=3 cm=[tex]\frac{3}{100}=0.03 m[/tex]
1 m=100 cm
We have to find the magnitude of the magnetic force on the electron if the electron velocity is directed as follows.
Magnetic field,B=[tex]\frac{2\mu_0 I}{4\pi d}[/tex]
[tex]\frac{\mu_0}{4\pi}=10^{-7}[/tex]
Substitute the values
[tex]B=\frac{2\times 10^{-7}\times 66}{(0.03)}=4.4\times 10^{-4} T[/tex]
q=[tex]1.6\times 10^{-19} C[/tex]
[tex]F=qVB=1.6\times 10^{-19}\times 1\times 10^7\times 4.4\times 10^{-4}=7.04\times 10^{-16} N[/tex]
