Question:

A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g = 10 m/s²) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire when it travels a vertical distance of 200 m is_______ mV.}

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Always be careful with unit conversions: $1 \text{ V} = 1000 \text{ mV}$ and $1 \text{ Gauss} = 10^{-4} \text{ Tesla}$.
Updated On: Feb 4, 2026
  • 0.2√10
  • 200√10
  • 2√10
  • 20√10
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
A conductor moving through a magnetic field experiences motional EMF given by $E = Blv$, where $v$ is the instantaneous velocity perpendicular to the field and length.
Step 2: Key Formula or Approach:
1. Velocity after falling height $h$: $v = \sqrt{2gh}$.
2. Induced EMF: $\varepsilon = Blv$.
3. Conversion: $1 \text{ Gauss} = 10^{-4} \text{ Tesla}$.
Step 3: Detailed Explanation:
Given $h = 200$ m, $g = 10$ m/s²: \[ v = \sqrt{2 \times 10 \times 200} = \sqrt{4000} = 20\sqrt{10} \text{ m/s} \] Magnetic field $B = 0.5 \text{ Gauss} = 0.5 \times 10^{-4}$ T. Length $l = 20$ m. \[ \varepsilon = (0.5 \times 10^{-4}) \times 20 \times (20\sqrt{10}) \] \[ \varepsilon = 10^{-4} \times 200\sqrt{10} = 2 \times 10^{-2}\sqrt{10} \text{ V} \] To convert to mV: \[ \varepsilon = 0.02\sqrt{10} \times 1000 = 20\sqrt{10} \text{ mV} \]
Step 4: Final Answer:
The induced EMF is 20√10 mV.
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