Question:
A 10 m long copper wire while remaining in the east-west horizontal direction is falling down with a speed of 5.0 m/s. If the horizontal component of the earthâs magnetic field \( B = 0.3 \times 10^{-4} \, \text{Wb/m}^2 \) the emf developed between the ends of the wires is:
A 10 m long copper wire while remaining in the east-west horizontal direction is falling down with a speed of 5.0 m/s. If the horizontal component of the earthâs magnetic field \( B = 0.3 \times 10^{-4} \, \text{Wb/m}^2 \) the emf developed between the ends of the wires is:
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When a conductor moves through a magnetic field, the induced emf depends on the velocity, magnetic field, and the length of the conductor.
Updated On: Apr 22, 2026
- 0.15 V
- 1.5 V
- 0.15 mV
- 1.5 mV
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The Correct Option is D
Solution and Explanation
Step 1: Use the formula for induced emf.
The induced emf \( \varepsilon \) in a conductor moving with velocity \( v \) perpendicular to a magnetic field \( B \) is given by: \[ \varepsilon = B \times v \times L \] where: - \( B = 0.3 \times 10^{-4} \, \text{Wb/m}^2 \) is the magnetic field, - \( v = 5.0 \, \text{m/s} \) is the velocity of the wire, - \( L = 10 \, \text{m} \) is the length of the wire.
Step 2: Calculate the induced emf.
Substitute the given values into the equation: \[ \varepsilon = 0.3 \times 10^{-4} \times 5.0 \times 10 = 1.5 \times 10^{-3} = 1.5 \, \text{mV} \]
Step 3: Conclusion.
The induced emf is 1.5 mV, which is option (4).
The induced emf \( \varepsilon \) in a conductor moving with velocity \( v \) perpendicular to a magnetic field \( B \) is given by: \[ \varepsilon = B \times v \times L \] where: - \( B = 0.3 \times 10^{-4} \, \text{Wb/m}^2 \) is the magnetic field, - \( v = 5.0 \, \text{m/s} \) is the velocity of the wire, - \( L = 10 \, \text{m} \) is the length of the wire.
Step 2: Calculate the induced emf.
Substitute the given values into the equation: \[ \varepsilon = 0.3 \times 10^{-4} \times 5.0 \times 10 = 1.5 \times 10^{-3} = 1.5 \, \text{mV} \]
Step 3: Conclusion.
The induced emf is 1.5 mV, which is option (4).
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