Question

A wire of length \(L\); having resistance \(R\) falls from a height \(l\) in earth's horizontal magnetic field \(B\). The current through the wire is ( \(g = \) acceleration due to gravity)

• A. \(\frac{BL \sqrt{2gL}}{R}\)
• B. \(\frac{BL \sqrt{2gL}}{R^2}\)
• C. \(\frac{2BLgL}{R^2}\)
• D. \(\frac{B^2L^2}{R}\)

Hint:

In such problems, the wire cuts magnetic field lines only if its motion is perpendicular to the field. Here horizontal field and vertical motion yield maximum induced emf. Use energy conservation to find velocity.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A wire of length \(L\) and resistance \(R\) falls from a height \(l\) (presumably \(l = L\) from the option form) in a uniform horizontal magnetic field \(B\). The wire falls vertically, cutting horizontal magnetic field lines, inducing an emf. We need the current.

Step 2: Key Formula or Approach:
Induced emf = \(B L v\), where \(v\) is the velocity of the wire. For a fall from height \(l\), velocity just before impact is \(v = \sqrt{2gl}\). Using Ohm’s law, current \(I = \frac{\text{emf}}{R}\).

Step 3: Detailed Explanation:
Assuming the wire falls from height \(l = L\) (as indicated by the option \(\sqrt{2gL}\)), the velocity is \(\sqrt{2gL}\). Then emf = \(B L \sqrt{2gL}\). Current = \(\frac{B L \sqrt{2gL}}{R}\). This matches option (A).

Step 4: Final Answer:
Option (A) is correct.