Problem

A $90 \, \mathrm{mm}$ length of wire moves with an upward velocity of $35 \, \mathrm{ms}^{-1}$ between the poles of a magnet. The magnetic field is $80 \, \mathrm{mT}$ directed to the right. The resistance in the wire is $5.00 \, \mathrm{m}\Omega$. Determine the magnitude and direction of the induced current.

Data

• Length of the wire: $L = 90 \, \mathrm{mm} = 0.09 \, \mathrm{m}$
• Magnetic field: $B = 80 \, \mathrm{mT} = 80 \times 10^{-3} \, \mathrm{T}$
• Velocity of the wire: $v = 35 \, \mathrm{ms}^{-1}$
• Resistance of the wire: $R = 5.00 \, \mathrm{m}\Omega = 5.00 \times 10^{-3} \, \Omega$

To find:

1. Induced current ($I$) - magnitude and direction.

Prerequisite Concepts

1. Motional e.m.f.:

$$\varepsilon = B L v$$

where:

• $B$ is the magnetic field,
• $L$ is the length of the wire,
• $v$ is the velocity of the wire.

2. Ohm’s Law for current:

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

where:

• $\varepsilon$ is the induced e.m.f.,
• $R$ is the resistance of the wire.

3. Fleming’s Right-Hand Rule determines the direction of the induced current.

Solution

Step 1: Calculate the Induced e.m.f.

1. Use the formula:

$$\varepsilon = B L v$$

2. Substitute the given values:

$$\varepsilon = (80 \times 10^{-3}) \cdot 0.09 \cdot 35$$

3. Simplify:

$$\varepsilon = 0.252 \, \mathrm{V}.$$

Step 2: Calculate the Induced Current

1. Use Ohm’s Law:

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

2. Substitute the values:

$$I = \frac{0.252}{5.00 \times 10^{-3}}.$$

3. Simplify:

$$I = 50.4 \, \mathrm{A}.$$

Step 3: Determine the Direction

1. Using Fleming’s Right-Hand Rule:
   • The thumb points in the direction of motion (upwards).
   • The forefinger points in the direction of the magnetic field (to the right).
   • The middle finger points in the direction of the induced current, which is into the plane of the paper.

Answer

1. Magnitude of induced current:

$$I = 50.4 \, \mathrm{A}.$$

2. Direction of the induced current: Into the plane of the paper.

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