Question

A long wire carrying I = 6.40 A of current makes two bends, as shown in the figure (Figure 1). The bent part of the wire passes through a uniform 0.270 T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude of the force that the magnetic field exerts on the wire.

Answer 1

The magnitude of the force that the magnetic field exerts on the wire is 1.728 N.

Step-by-step explanation:

To calculate the magnitude of the force that the magnetic field exerts on the wire, we can use the equation F = I−ǸB sin, where F is the force, I is the current, B is the magnetic field, and is the angle between I and B. In this case, the current is 6.40 A and the magnetic field is 0.270 T. Since the wire makes two bends, we'll need to consider the force for each bend separately.

For the first bend, the angle between the current and magnetic field is 180°, so sin = 0. The force on this bend is therefore 0.

For the second bend, the angle between the current and magnetic field is 90°, so sin = 1. Plugging in the values, we get F = 6.40 A − 0.270 T × 1 = 1.728 N.