Induced EMF and Magnetic Field in a Slide Wire Generator

Q11. A slide wire generator with R = 1000 ohms contains a moveable wire with length 0.05 meters. The current induced in the wire is 0.005 A. If we pull the wire to the right at 400 m/s, what EMF ‘𝜀’ is induced in the loop and what is the magnetic field magnitude ‘B’?

Background

Topic: Electromagnetic Induction (Faraday's Law, Motional EMF)

This question tests your understanding of how a moving conductor in a magnetic field induces an electromotive force (EMF), and how to relate that to the induced current and the magnetic field strength.

Key Terms and Formulas

• Motional EMF: When a conductor of length $L$ moves with velocity $v$ perpendicular to a magnetic field $B$, the induced EMF is given by:

  $\varepsilon = vLB$

• Ohm's Law: The induced current $I$ in the circuit is related to the EMF and resistance $R$ by:

  $I = \frac{\varepsilon}{R}$

Step-by-Step Guidance

1. Start by identifying the known values: $R = 1000\ \Omega$, $L = 0.05\ \text{m}$, $I = 0.005\ \text{A}$, $v = 400\ \text{m/s}$.

2. Use Ohm's Law to relate the induced EMF to the current and resistance:

   $\varepsilon = IR$

3. Calculate the induced EMF using the values for $I$ and $R$ (but do not compute the final value yet).

4. Set the motional EMF formula equal to the value from Ohm's Law: $vLB = IR$.

5. Rearrange the equation to solve for the magnetic field $B$:

   $B = \frac{IR}{vL}$

Try solving on your own before revealing the answer!

Final Answer: 5.0 V and 0.25 T

Using $\varepsilon = IR = (0.005)(1000) = 5.0\ \text{V}$ and $B = \frac{5.0}{400 \times 0.05} = 0.25\ \text{T}$.

The EMF is found from Ohm's Law, and the magnetic field is calculated by rearranging the motional EMF formula.