Question

Two identical circular, wire loops 35.0 cm in diameter each carry a current of 2.80 A in the same direction. These loops are parallel to each other and are 24.0 cm apart. Line ab is normal to the plane of the loops and passes through their centers. A proton is fired at 2600 m/s perpendicular to line ab from a point midway between the centers of the loops.

Find the magnitude of the magnetic force these loops exert on the proton just after it is fired.

Answer 1

The answer is "
`4659.2 * 10^(-24) \ N`"

Step-by-step explanation:

The magnetic field at ehe mid point of the coils is,

`\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{(3)/(2)}}\\\\`

Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.

`\to B=\frac{(4\pi* 10^(-7))(2.80\ A) ((0.35)/(2))^2}{( ((0.35)/(2))^2+ ((0.24)/(2))^2)^{(3)/(2)}}\\\\`

`=\frac{(12.56 * 10^(-7))(2.80\ A) * 0.030625}{( 0.030625+ 0.0144)^{(3)/(2)}}\\\\=( 1.07702 * 10^(-7) )/(0.0095538976)\\\\=112.730955 * 10^(-7)\\\\=1.12* 10^(-5)\ \ T\\`

Calculating the force experienced through the protons:

`F=qvB=(1.6 * 10^(-19)) (2600)(1.12 * 10^(-5))= 4659.2 * 10^(-24)\ N`