Question

A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B field (B is constant in time) and then back into a field free region to the left. The self inductance of the loop is negligible.

A square, single-turn wire coil L = 1.20 cm on a side is placed inside a solenoid that has a circular cross section of radius r = 2.85 cm, as shown in the figure below. The solenoid is 22.3 cm long and wound with 147 turns of wire. If the current in the solenoid is 2.63 A, find the flux through the coil. If the current in the solenoid is reduced to zero in 3.03 s, calculate the magnitude of the average induced emf in the coil.

When entering the field the coil experiences a magnetic force to the left.
When leaving the field the coil experiences a magnetic force to the right.
Upon entering the field, a clockwise current flows in the loop.
Upon leaving the field, a clockwise current flows in the loop.

Answer 1

As the square enters the field, the magnetic flux is increasinginto the page, meaning that a current will be created to fight thatincrease. So, while entering, the current is counterclockwise (B)and the coil experiences a force to the right. When the square isentirely in the field there is no change in magnetic flux so thereis no induced current (E). When leaving the field, the magneticflux is decreasing into the board, so to fight it, we need anincrease into the board, so the current is clockwise and the forceit experiences is to the left.

Upon entering the field, a clockwise current flows in the loop.

Answer 2

1

True
Current directions in the top and the bottom sides of the loop are in opposite directions. Both are in the same magnetic field. Therefore, forces on them are equal and opposite and cancel each other. The right side of the loop is outside the magnetic field and, therefore, experiences no magnetic force.
So, the net force on the loop is the force on the left side. The current in the loop is clockwise. For this, the current in the left side is from bottom to top.
And magnetic field is out of the plane of figure. The direction of force is in the direction of cross product of direction of current and direction of field. This direction is towards right.

2

False
When inside the field, then magnetic flux through the coil is out of plane of figure. When leaving the field, the flux decreases. This means that change in flux is into the plane of figure.
Therefore, the current in the loop is such that the flux produced by it is out of the plane of figure. For that the magnetic field produced by the current should be out of the plane of figure. This is possible when the current is anticlockwise.
We found in (2) that clockwise current produces magnetic force towards right. By similar reasoning, anticlockwise current will produce magnetic force to the left.

3

True
Before entering the field, the magnetic flux through the coil is zero. Upon entering the field, the flux is out of the plane of figure because magnetic field is out of the plane of figure. So, the change in flux is out of the plane of figure.
Therefore, the current in the loop is such that the flux produced by it is into the plane of figure. For that the magnetic field produced by the current should be into the plane of figure. This is possible when the current is clockwise.

4

true

Upon leaving the field, a clockwise current flows in the loop.