Question

A proton of charge $e$ moving at speed $v_0$ is placed midway between two parallel wires a distance $a$ apart, each carrying current $I$ in opposite directions.
The force on the proton is:

• A. $0$
• B. $e{v}_0\frac{{\mu }_{0}I}{2\pi a}$
• C. $e{v}_0\frac{{\mu }_{0}I}{2\pi \left(2a\right)}$
• D. $e{v}_0\frac{{\mu }_{0}I}{2\pi \frac{a}{2}}$
• E. Unable to be determined

Answer 1

The correct option is A $0$

In fact proton is under the effect of two equal magnetic fields due to two wires.

Let $a$ is the distance between two wires.

The magnetic field due to first wire carrying a current $I$ at a distance $a/2$ (midway) is given by:

$B_1=\frac{{\mu }_0}{2\pi }.\frac{I}{a/2}$

Magnetic field due to second wire carrying a current $I$ at a distance $a/2$ (midway) is given by:

$B_2=\frac{{\mu }_0}{2\pi }.\frac{I}{a/2}$

As the direction of current is opposite in two wires therefore by Right hand thumb rule, both the magnetic fields will be in the same direction, therefore force on proton due to two fields will be the sum of forces due to two fields.

$F=e{v}_0{B}_1-e{v}_0{B}_2=e{v}_0\frac{{\mu }_0}{2\pi }.\frac{I}{a/2}-e{v}_0\frac{{\mu }_0}{2\pi }.\frac{I}{a/2}$

or $F=0$