Problem

A rectangular wire loop with sides $8 \, \mathrm{cm}$ and $2 \, \mathrm{cm}$ and a small cut moves through a uniform magnetic field of magnitude $0.3 \, \mathrm{T}$ directed normal to the plane of the loop. Determine:
(a) The e.m.f. induced when the velocity of the loop ($v = 1 \, \mathrm{cm/s}$) is normal to the longer side.
(b) The e.m.f. induced when the velocity of the loop is normal to the shorter side.
(c) The time taken in both cases.

Data

• Length of the loop: $L = 8 \, \mathrm{cm} = 0.08 \, \mathrm{m}$
• Width of the loop: $W = 2 \, \mathrm{cm} = 0.02 \, \mathrm{m}$
• Magnetic field: $B = 0.3 \, \mathrm{T}$
• Velocity of the loop: $v = 1 \, \mathrm{cm/s} = 0.01 \, \mathrm{m/s}$

To find:

1. E.m.f. induced when moving normal to the longer side ($\varepsilon_L$).
2. E.m.f. induced when moving normal to the shorter side ($\varepsilon_S$).
3. Time taken in both cases ($\Delta t$).

Prerequisite Concepts

1. Motional e.m.f. formula:

$$\varepsilon = B L v$$

where $L$ is the length of the side of the loop moving through the field.

2. Time taken to move across the field:

$$t = \frac{S}{v},$$

where $S$ is the distance moved.

Solution

Part (a): E.m.f. Induced in the Longer Side

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

$$\varepsilon_L = B L v.$$

2. Substituting values:

$$\varepsilon_L = 0.3 \cdot 0.08 \cdot 0.01.$$

3. Simplify:

$$\varepsilon_L = 2.4 \times 10^{-4} \, \mathrm{V}.$$

Part (b): E.m.f. Induced in the Shorter Side

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

$$\varepsilon_S = B W v.$$

2. Substituting values:

$$\varepsilon_S = 0.3 \cdot 0.02 \cdot 0.01.$$

3. Simplify:

$$\varepsilon_S = 0.6 \times 10^{-4} \, \mathrm{V}.$$

Part (c): Time Taken

1. For the longer side:

$$t_L = \frac{W}{v}.$$

2. Substituting values:

$$t_L = \frac{0.02}{0.01}.$$

3. Simplify:

$$t_L = 2 \, \mathrm{s}.$$

Answer

1. Induced e.m.f. when moving normal to the longer side:

$$\varepsilon_L = 2.4 \times 10^{-4} \, \mathrm{V}.$$

2. Induced e.m.f. when moving normal to the shorter side:

$$\varepsilon_S = 0.6 \times 10^{-4} \, \mathrm{V}.$$

3. Time taken in both cases:

$$t_L = 2 \, \mathrm{s}.$$

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