Problem

A horizontal straight wire $10 \, \mathrm{m}$ long, extending from east to west, is falling with a speed of $5.0 \, \mathrm{ms}^{-1}$ at right angles to the horizontal component of the Earth’s magnetic field $0.30 \times 10^{-4} \, \mathrm{Wb/m}^{2}$.

(a) What is the instantaneous value of the e.m.f. induced in the wire?
(b) What is the direction of the e.m.f.?
(c) Which end of the wire is at the higher electrical potential?

Data

• Length of the wire: $L = 10 \, \mathrm{m}$
• Magnetic field: $B = 0.30 \times 10^{-4} \, \mathrm{Wb/m}^{2}$
• Speed of the wire: $v = 5.0 \, \mathrm{ms}^{-1}$

To find:

1. Induced e.m.f. $\varepsilon$.
2. Direction of the e.m.f.
3. End at higher potential.

Prerequisite Concepts

1. Motional e.m.f. formula:

$$\varepsilon = B L v$$

2. Fleming’s Right-Hand Rule determines the direction of the induced current and e.m.f.
3. Higher potential end is determined by the flow of current, with the direction of e.m.f. indicating higher to lower potential.

Solution

Part (a): Induced e.m.f.

1. Using the formula for motional e.m.f.:

$$\varepsilon = B L v$$

2. Substituting the values:

$$\varepsilon = (0.30 \times 10^{-4}) \cdot 10 \cdot 5.0$$

3. Simplify:

$$\varepsilon = 1.5 \times 10^{-3} \, \mathrm{V} = 1.5 \, \mathrm{mV}.$$

Part (b): Direction of the e.m.f.

1. Applying Fleming’s Right-Hand Rule:
   • Stretch the thumb, forefinger, and middle finger of the right hand perpendicular to each other.
   • The forefinger points in the direction of the magnetic field (downwards, into the plane).
   • The thumb points in the direction of motion (falling downwards).
   • The middle finger indicates the direction of the induced current, which is from west to east.

Part (c): End at Higher Potential

1. Since the current flows from west to east, the western end of the wire is at a higher potential.

Answer

1. Induced e.m.f.:

$$\varepsilon = 1.5 \, \mathrm{mV}.$$

2. Direction of e.m.f.: From west to east.
3. The western end of the wire is at a higher potential.

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