The question asks what would happen if a coil made from thinner wire but with the same number of turns and coil diameter is used. Let's break it down:

• Thinner wire: This means the wire has a smaller cross-sectional area, which leads to an increase in resistance according to the formula $R = \frac{\rho L}{A}$, where $A$ is the cross-sectional area.
• Same number of turns and diameter: This keeps the coil geometry constant.

Impact:

• Increased resistance: Thinner wire has higher resistance, reducing the current flowing through the coil for the same applied voltage (Ohm’s Law: $V = IR$).
• Torque: Torque in a coil (e.g., in motors) depends on the current and magnetic field. Since the current decreases due to higher resistance, the torque produced will also reduce.

Correct answer:

• Both B and C: The torque produced will decrease, and the current flowing through the coil will also reduce.

Let me know if you'd like further details or clarification!

Here are 5 questions related to this concept:

1. How does the cross-sectional area of a wire affect its resistance?
2. What is the relationship between current and torque in a coil?
3. How does increasing the resistance affect the current in a circuit with constant voltage?
4. What other factors, aside from resistance, can influence torque in an electromagnetic coil?
5. How is power loss in a coil affected by the resistance of the wire?

Tip: In electromagnetism, the performance of a coil depends heavily on the wire's resistance, which influences the current flow and the resulting magnetic field.