CBSE · Class 12 · Physics · Electromagnetic InductionA rectangular wire loop of sides $8\text{ cm}$ and $2\text{ cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3\text{ T}$ directed normal to the loop. What is the electromotive force (emf) developed across the cut if the velocity of the loop is $1\text{ cm/s}$ in a direction normal to the: (a) longer side of the loop? (b) shorter side of the loop?\nFor how long does the induced voltage last in each case?

Step-by-Step Solution:

Given Data:

• Length of longer side, $l = 8\text{ cm} = 0.08\text{ m}$
• Length of shorter side, $b = 2\text{ cm} = 0.02\text{ m}$
• Magnetic field strength, $B = 0.3\text{ T}$
• Velocity of the loop, $v = 1\text{ cm/s} = 0.01\text{ m/s}$

Case (a): Velocity is normal to the longer side ($l = 0.08\text{ m}$)

1. Calculation of Induced EMF ($\varepsilon$): When the loop moves out with velocity perpendicular to its longer side, the longer side cuts the magnetic field lines. $$\varepsilon = B \cdot l \cdot v$$ $$\varepsilon = 0.3\text{ T} \times 0.08\text{ m} \times 0.01\text{ m/s}$$ $$\varepsilon = 2.4 \times 10^{-4}\text{ V} = 0.24\text{ mV}$$

2. Calculation of Duration ($t_1$): The loop has to travel a distance equal to its shorter side ($b = 0.02\text{ m}$) to completely exit the magnetic field. $$t_1 = \frac{\text{Distance}}{\text{Velocity}} = \frac{b}{v}$$ $$t_1 = \frac{0.02\text{ m}}{0.01\text{ m/s}} = 2\text{ s}$$

Case (b): Velocity is normal to the shorter side ($b = 0.02\text{ m}$)

1. Calculation of Induced EMF ($\varepsilon$): When the loop moves out with velocity perpendicular to its shorter side, the shorter side cuts the magnetic field lines. $$\varepsilon = B \cdot b \cdot v$$ $$\varepsilon = 0.3\text{ T} \times 0.02\text{ m} \times 0.01\text{ m/s}$$ $$\varepsilon = 0.6 \times 10^{-4}\text{ V} = 0.06\text{ mV}$$

2. Calculation of Duration ($t_2$): The loop has to travel a distance equal to its longer side ($l = 0.08\text{ m}$) to completely exit the magnetic field. $$t_2 = \frac{\text{Distance}}{\text{Velocity}} = \frac{l}{v}$$ $$t_2 = \frac{0.08\text{ m}}{0.01\text{ m/s}} = 8\text{ s}$$

Final Answer:

• (a) Induced EMF = $2.4 \times 10^{-4}\text{ V}$ (or $0.24\text{ mV}$), lasting for $2\text{ seconds}$.
• (b) Induced EMF = $0.6 \times 10^{-4}\text{ V}$ (or $0.06\text{ mV}$), lasting for $8\text{ seconds}$.