Magnetic forces aon currents and magnetic induction

[abc123]

HELP ME PLEASE!

Given B=0.80 T, a square coil with a single loop with length of each side of 20 cm, a current of 5.0 A where there is an external constant uniform magnetic field, what are the size and the direction of the magnetic force on just the lower section of the loop due to the 0.80 T field?

 

[FrogPad]

The answer is 5.




seriously, have you tried anything?

 

[kyle8921]

The magnetic force on a current-carrying loop is given by the equation F = IABsinθ, where I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the current and the magnetic field. In this case, the current is 5.0 A, the area of the loop is (0.2 m)^2 = 0.04 m^2, and the magnetic field strength is 0.80 T. Since the loop is oriented perpendicular to the magnetic field, θ = 90°, and the sine of 90° is equal to 1. Plugging in these values, we can calculate the magnetic force on the lower section of the loop to be:

F = (5.0 A)(0.04 m^2)(0.80 T)(1) = 0.16 N

The direction of this force can be determined using the right-hand rule, where the thumb points in the direction of the current, the fingers point in the direction of the magnetic field, and the palm indicates the direction of the force. In this case, the force will be directed out of the page.

It is important to note that this calculation only applies to the lower section of the loop, as the magnetic force on the upper section would be equal in magnitude but opposite in direction, resulting in a net force of zero on the entire loop. Additionally, this calculation assumes that the loop is in a uniform magnetic field, meaning that the magnetic field is the same at all points within the loop. If the magnetic field were not uniform, the force on the loop would vary at different points.