Question

A metal rod of length 50 cm is held vertically and moved with a velocity in a magnetic field at the place of 0.4 G. The emf induced across the ends of the rod is:

• A. 0.1 mV
• B. 0.2 mV
• C. 0.8 mV
• D. 1.6 mV

Hint:

For motional emf, use the formula \( \text{emf} = B \ell v \sin \theta \). Ensure the velocity, magnetic field, and rod’s orientation are perpendicular for maximum emf. Convert units carefully, especially magnetic field from Gauss to Tesla.

Answer 1

The Correct Option is C

Step 1: Recall the formula for motional electromotive force (emf). The emf induced in a conductor of length \( \ell \) moving with velocity \( v \) through a magnetic field \( B \) is given by \( \text{emf} = B \ell v \sin \theta \), where \( \theta \) is the angle between the velocity vector and the magnetic field vector.

Step 2: Convert all given physical quantities to the International System of Units (SI).

• Length of the rod: \( \ell = 50 \, \text{cm} = 0.5 \, \text{m} \).
• Magnetic field strength: \( B = 0.4 \, \text{G} = 0.4 \times 10^{-4} \, \text{T} = 4 \times 10^{-5} \, \text{T} \). Note that 1 Gauss (G) is equivalent to \( 10^{-4} \) Tesla (T).
• Angle \( \theta \): Given that the rod is held vertically and moved, it is typical in such problems to assume the magnetic field is horizontal and the velocity is perpendicular to both the rod and the field. Therefore, \( \theta = 90^\circ \), and \( \sin \theta = 1 \).

Step 3: Determine a plausible velocity \( v \) by examining the provided options. Since the velocity is not explicitly stated, we will use the motional emf formula and work backward from the answer choices. The formula simplifies to \( \text{emf} = B \ell v \) under the assumption \( \sin \theta = 1 \). Substituting the known SI values: \[ \text{emf} = (4 \times 10^{-5} \, \text{T}) \times (0.5 \, \text{m}) \times v = (2 \times 10^{-5}) v \, \text{V} \] The answer options are in millivolts (mV). Convert the emf to millivolts: \[ \text{emf (in mV)} = (2 \times 10^{-5}) v \times 1000 = 0.02 v \, \text{mV} \] Let's test option (C), which suggests an emf of 0.8 mV: \[ 0.8 \, \text{mV} = 0.02 v \, \text{mV} \] Solving for \( v \): \[ v = \frac{0.8}{0.02} = 40 \, \text{m/s} \] A velocity of 40 m/s is a physically reasonable value for this type of problem, making it a likely candidate.

Step 4: Calculate the motional emf using the derived velocity. Using the calculated velocity \( v = 40 \, \text{m/s} \) and the formula \( \text{emf} = (2 \times 10^{-5}) v \): \[ \text{emf} = (2 \times 10^{-5}) \times 40 = 0.0008 \, \text{V} \] Converting this to millivolts: \[ 0.0008 \, \text{V} = 0.8 \, \text{mV} \] This calculated emf matches option (C).

Step 5: Review the assumptions made. The assumption that the velocity vector and the magnetic field vector are perpendicular (\( \sin \theta = 1 \)) is justified by the typical setup of such problems where a vertical rod moves horizontally in a horizontal magnetic field. If the angle \( \theta \) were different, the \( \sin \theta \) factor would alter the resulting emf. However, the precise match with option (C) strongly supports the validity of our assumptions regarding the orientation of the field and velocity.