Answer to Question #310227 in Electricity and Magnetism for Manisha

Question

Two long parallel wires, 12.0 cm apart are carrying current 1.8 A and 3.5 A respectively in the same directions. The magnetic field at the midpoint between both wires is 5.0 X 10^-8 T out of the page. If a proton moves with a velocity of 2.8 X 10⁵ m s-1 at the midpoint between the wires, determine the magnetic force acting on the proton.

Answer 1

Magnetic force

$B_m=\frac{\mu_0i_1}{2\pi d}-\frac{\mu_0i_2}{2\pi d}$

$$B_m=\frac{\mu_0\times3.5}{2\pi \times0.06}-\frac{\mu_0\times1.8}{2\pi \times0.06}=5.00\times10^{-8}T(approximately)$$

Magnetic force

$F=qvB$

$$F=1.6\times10^{-19}\times2.8\times10^5\times5\times10^{-8}=2.24\times10^{-21}N$$