Figure 6 shows a wire in a magnetic field - AQA - GCSE Physics Combined Science - Question 4 - 2019 - Paper 2

1. Reverse the direction of the current.

2. Reverse the direction of the magnetic field.

To find the current, we can use the formula: $F = B I L$ where:

• $F$ is the force on the wire (0.072 N)
• $B$ is the magnetic flux density (360 mT or 0.360 T)
• $I$ is the current in the wire
• $L$ is the length of the wire (0.050 m)

Rearranging the formula to solve for $I$ gives us: $I = \frac{F}{B \cdot L}$ Substituting the values: $I = \frac{0.072}{0.360 \times 0.050}$ Calculating this: $I = \frac{0.072}{0.018} = 4.0 A$

When there is a current in the coil, a magnetic field is created by the permanent magnets. As the current flows, it interacts with the magnetic field, resulting in forces acting on the different sides of the coil. These forces act in opposite directions, causing the coil to rotate. The split-ring commutator helps to switch the direction of the current every half rotation, which maintains the rotation of the coil in the same direction.