Question

A metal rod of length 0.5 m moves at $3~ms^{-1}$ perpendicular to a magnetic field of 0.2 T. The induced emf is ________.

• A. 0.1 V
• B. 0.2 V
• C. 0.3 V
• D. 0.4 V
• E. 0.6 V

Hint:

Induced EMF = (Magnetic Field) $\times$ (Velocity) $\times$ (Length).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This problem deals with motional electromotive force (EMF). When a conductor moves through a magnetic field, the magnetic force on the free charge carriers within the conductor causes them to accumulate at the ends of the conductor. This separation of charge creates an electric field and a potential difference, which is the induced EMF.
Step 2: Key Formula or Approach
The magnitude of the motional EMF (\(\mathcal{E}\)) induced in a straight conductor of length \(L\) moving with velocity \(v\) in a uniform magnetic field \(B\) is given by:
\[ \mathcal{E} = B L v \sin\theta \] where \(\theta\) is the angle between the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\). The formula assumes that the length of the rod, its velocity, and the magnetic field are mutually perpendicular.
The problem states the rod "moves with its length perpendicular to a uniform magnetic field". This implies the velocity is also perpendicular to the field for maximum EMF. Let's assume the velocity vector is perpendicular to both the rod's length and the magnetic field. In this standard configuration, the formula simplifies to:
\[ \mathcal{E} = B L v \] Step 3: Detailed Explanation
1. List the given values.
- Magnetic field, \(B = 0.2 \text{ T}\)
- Length of the rod, \(L = 0.5 \text{ m}\)
- Velocity, \(v = 3 \text{ ms}^{-1}\)
The problem describes a situation where B, L, and v are mutually perpendicular, which is the condition for maximum EMF.
2. Calculate the induced EMF.
Using the formula \(\mathcal{E} = B L v\):
\[ \mathcal{E} = (0.2 \text{ T}) \times (0.5 \text{ m}) \times (3 \text{ ms}^{-1}) \] \[ \mathcal{E} = (0.2 \times 0.5) \times 3 \] \[ \mathcal{E} = (0.1) \times 3 \] \[ \mathcal{E} = 0.3 \text{ V} \] Step 4: Final Answer
The induced emf in the rod is 0.3 V.