Above the ground, a 1 m long wire is suspended horizontally perpendicular to a line running from its center to a nearby observer. A uniform magnetic field of 1.8 Tesla permeates the volume occupied by this wire in the upward direction.

a) Ifa current of 5x10-2 A is passed thru the wire, what is the force on the wire?
b) What is its direction?

a) 0.09N b) positive x direction

Explanation:

Force on a conductor carrying current in a magnetic field can be expressed as;

F = BILsin(theta) where

F is the force on the conductor (wire)

B is the uniform magnetic field I'm Tesla = 1.8Tesla

I is the current in the wire = 5×10^-2A

L is the length of the wire = 1m

theta is the angle that the conductor make with the magnetic field = 90° (since the wire in the horizontal direction is perpendicular to the field acting upwards)

Substituting this value in the formula to get F we have;

F = 1.8×5×10^-2×1 × sin90°

F = 0.09N

The force on the wire is 0.09N

b) The direction of the force is in the positive x direction since the wire acts horizontally to the magnetic field.

truck y has a greater velocity than truck x.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards