Question

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 \(\Omega\) then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is _____ N.

• A. \(5.7 \times 10^{-2}\)
• B. \(7.5 \times 10^{-3}\)
• C. \(7.5 \times 10^{-2}\)
• D. \(5.7 \times 10^{-3}\)

Hint:

An alternative approach is to use the principle of energy conservation. The mechanical power supplied by the external force must equal the electrical power dissipated as heat in the resistor.
Mechanical Power \(P_{\text{mech}} = F_{\text{app}} \cdot v\).
Electrical Power \(P_{\text{elec}} = I^2 R = \frac{\mathcal{E}^2}{R} = \frac{(BLv)^2}{R}\).
Equating them: \(F_{\text{app}} v = \frac{B^2 L^2 v^2}{R} \implies F_{\text{app}} = \frac{B^2 L^2 v}{R}\). This gives the answer in one step.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A conducting rod moving in a magnetic field as part of a closed circuit will experience motional EMF. This EMF drives a current, and the current-carrying rod in the magnetic field experiences a magnetic (Lorentz) force. To maintain constant velocity, an external applied force equal and opposite to this magnetic force is required.
Step 2: Key Formula or Approach:
1. Motional EMF: \( \mathcal{E} = B L v \), where B, L, and v are mutually perpendicular. 2. Ohm's Law: \( I = \frac{\mathcal{E}}{R} \). 3. Magnetic Force on the rod: \( F_m = I L B \). 4. Condition for constant velocity: Applied Force \( F_{\text{app}} = F_m \).
Step 3: Detailed Explanation:
Given values:
Length of the rod, \(L = 1\) m.
Magnetic field, \(B = 0.10\) T.
Total resistance, \(R = 2 \, \Omega\).
Constant speed, \(v = 1.5\) m/s.
1. Calculate the induced motional EMF (\(\mathcal{E}\)):
\[ \mathcal{E} = B L v = (0.10 \, \text{T}) \times (1 \, \text{m}) \times (1.5 \, \text{m/s}) = 0.15 \, \text{V} \] 2. Calculate the induced current (I):
\[ I = \frac{\mathcal{E}}{R} = \frac{0.15 \, \text{V}}{2 \, \Omega} = 0.075 \, \text{A} \] 3. Calculate the magnetic force (\(F_m\)) on the rod:
This force will oppose the motion (by Lenz's law). Its magnitude is:
\[ F_m = I L B = (0.075 \, \text{A}) \times (1 \, \text{m}) \times (0.10 \, \text{T}) = 0.0075 \, \text{N} \] 4. Determine the required applied force (\(F_{\text{app}}\)):
To move the rod at a constant speed, the net force must be zero. Therefore, the applied force must be equal in magnitude and opposite in direction to the magnetic force.
\[ F_{\text{app}} = F_m = 0.0075 \, \text{N} \] In scientific notation, this is \( 7.5 \times 10^{-3} \) N.
Step 4: Final Answer:
The force needed to move the rod is \( 7.5 \times 10^{-3} \) N. This corresponds to option (B).