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 :
[tex]B=\dfrac{\mu_o Ir^2}{(r^2+x^2)^{3/2}}[/tex]
[tex]B=\dfrac{4\pi \times 10^{-7}\times 4.5\times (0.24)^2}{((0.24)^2+(0.3/2)^2)^{3/2}}[/tex]
[tex]B=1.43\times 10^{-5}\ T[/tex]
Let F is the magnetic force these loops exert on the proton just after it is fired. It is given by :
[tex]F=q\times v\times B[/tex]
[tex]F=1.6\times 10^{-19}\times 2600\times 1.43\times 10^{-5}[/tex]
[tex]F=5.94\times 10^{-21}\ N[/tex]
So, the magnetic force these loops exert on the proton is [tex]5.94\times 10^{-21}\ N[/tex]. Hence, this is the required solution.
