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

Answer:

The answer is "$4659.2 \times 10^{-24} \ N$"

Explanation:

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

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

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.

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

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

Calculating the force experienced through the protons:

$F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N$