Question

Two identical circular, wire loops 48.0 cm in diameter each carry a current of 4.50 A in the same direction. These loops are parallel to each other and are 30.0 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. Part A Find the magnitude of the magnetic force these loops exert on the proton just after it is fired. FF = nothing N

Answer 1

Explanation:

It is given that,

Diameter of loops, d = 48 cm = 0.48 m

Radius of the loop, r = 0.24 m

Current carried in the loop, I = 4.5 A

Distance between loops, x = 30 cm = 0.3 m

Speed of the proton, v = 2600 m/s

We know that the magnetic field at the midway of the coils is given by :

$B=\dfrac{\mu_o Ir^2}{(r^2+x^2)^{3/2}}$

$B=\dfrac{4\pi \times 10^{-7}\times 4.5\times (0.24)^2}{((0.24)^2+(0.3/2)^2)^{3/2}}$

$B=1.43\times 10^{-5}\ T$

Let F is the magnetic force these loops exert on the proton just after it is fired. It is given by :

$F=q\times v\times B$

$F=1.6\times 10^{-19}\times 2600\times 1.43\times 10^{-5}$

$F=5.94\times 10^{-21}\ N$

So, the magnetic force these loops exert on the proton is $5.94\times 10^{-21}\ N$. Hence, this is the required solution.