The force on the rectangular loop of wire in the magnetic field in the figure below can be used to measure field strength. The field is uniform, and the plane of the loop is perpendicular to the field. Justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop.

Question

The force on the rectangular loop of wire in the magnetic field in the figure below can be used to measure field strength. The field is uniform, and the plane of the loop is perpendicular to the field. Justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop.

If a current of 5.70 A is used, what is the force per tesla on the 32.0–cm–wide loop? (You do not need to enter any units.)

Answer 1

To justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop, we can analyze the properties of magnetic fields and current-carrying loops.

The force on a current-carrying wire in a magnetic field is given by the formula F = BIL, where F is the force, B is the magnetic field strength, I is the current, and L is the length of the wire segment that is in the magnetic field.

In the case of a rectangular loop, each side of the loop will experience an equal and opposite force due to the symmetry of the loop. The forces on the sides that are parallel to the magnetic field will be in the same direction and add up to zero, while the forces on the sides perpendicular to the magnetic field will be equal in magnitude and opposite in direction.

Therefore, regardless of how much of the loop is in the field, the forces on the sides will cancel each other out, resulting in a net force of zero on the loop. This is because the forces on adjacent sides are always equal and opposite.

Now, to calculate the force per tesla on the 32.0-cm-wide loop when a current of 5.70 A is used, we can use the formula mentioned earlier: F = BIL.

In this case, the width of the loop, which is perpendicular to the magnetic field, is given as 32.0 cm, or 0.32 m (since 1 cm = 0.01 m).

The current (I) is given as 5.70 A.

To find the force per tesla, we need to divide the force by the magnetic field strength (B). So the formula can be rearranged as B = F / (IL).

Since the net force on the loop is zero, the force per tesla is also zero. Therefore, the force per tesla on the 32.0-cm-wide loop is zero.

Note: It is important to double-check the units and make sure they are consistent throughout the calculations.