1)A current carrying loop of width a and length b is placed near a current carrying wire. How does the net force on the loop compare to the net force on a single wire segment of length a carrying the same amount of current placed at the same distance from the wire?

The net force on a current carrying loop is equal to zero. This is because the the loop experiences equal but opposite forces on the opposite sides of the loop due to the magnetic field created by the current flowing through the wire. On adding the forces acting algebraically. The sum up to zero. The second wire on the other hand has a net force acting on it equal in magnitude to ILBSin(theta).

So the wire experiences a net force compared to the loop that experiences no net force since all forces cancel out.

Explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s