Find the force due to a 0.35 T magnetic field directed it the east on a straight

Question

Find the force due to a 0.35 T magnetic field directed it the east on a straight wire segment of length 2.15 m carrying a current of 7.00 A to the south.
A. Compute the force on a proton moving to the west at a speed of 4.89 x 10^6 m/s , due to a downward directed magnetic field of intesity 5.20 x 10^-5 T. Find the force due to a 0.35 T magnetic field directed it the east on a straight wire segment of length 2.15 m carrying a current of 7.00 A to the south.
A. Compute the force on a proton moving to the west at a speed of 4.89 x 10^6 m/s , due to a downward directed magnetic field of intesity 5.20 x 10^-5 T.
A. Compute the force on a proton moving to the west at a speed of 4.89 x 10^6 m/s , due to a downward directed magnetic field of intesity 5.20 x 10^-5 T.

Explanation / Answer

The magnetic force is given by;

FB=I L B sin q

The strength of magnetic field is 0.35 T, length of line segment is 2.15 m and magnitude of current is 7.00 A.

The angle subtended by direction of current and magnetic field is 900( As magnetic field is in east direction and current in south direction).

So,

FB=I L B sin q

FB=7 A* 2.15 m* 0.35 T sin 900

FB=7 A* 2.15 m* 0.35 T *1

FB=5.2675 N

The magnetic force on a moving charge is given by;

F= Charge*Velocity*Magnetic field

So, charge of a proton is given by;

F= charge*velcity*magnetic field

F=1.6*10^-19 c*4.80*10^6 m/s*5.20*10^-5 T

F=39.936*10-18 N

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