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Naya [18.7K]
3 years ago
10

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 par

allel 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
Physics
1 answer:
maw [93]3 years ago
7 0

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.

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