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

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 :

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

So, the magnetic force these loops exert on the proton is . Hence, this is the required solution.

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.