Question

A wire of length $1\ \text{m}$ is moving at a speed of $2\ \text{m/s}$ perpendicular to a homogeneous magnetic field of $0.5\ \text{T}$. If the ends of the wire are joined to a resistance of $6\ \Omega$, the rate at which work is being done to keep the wire moving at that speed is

• A. $\frac{1}{3}\ \text{W}$
• B. $\frac{1}{6}\ \text{W}$
• C. $\frac{1}{12}\ \text{W}$
• D. $1\ \text{W}$

Hint:

To ensure accuracy, you can compute this problem using two completely independent paths: the electrical dissipation path ($P = e^2/R$) and the mechanical mechanical path ($P = Fv$). Since both methods yield exactly $\frac{1}{6}\ \text{W}$, you can be fully confident in your answer during a test!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A straight conducting wire moves through a uniform magnetic field, experiencing motional electromotive force (emf). The wire forms a closed circuit across a resistor. We need to find the mechanical power input (rate of doing work) required to sustain this constant velocity against the opposing magnetic braking force.

Step 2: Key Formula or Approach:
1. The motional emf ($e$) induced across a conductor moving perpendicularly through a magnetic field is: $$e = B \cdot l \cdot v$$ 2. The induced current ($I$) circulating through the closed loop resistance $R$ is given by Ohm's law: $$I = \frac{e}{R}$$ 3. By the conservation of energy, the mechanical power input ($P$) needed to maintain steady motion equals the electrical power dissipated as Joule heat inside the resistor: $$P = \frac{e^2}{R}$$

Step 3: Detailed Explanation:
Let's list the values provided in the problem: Length of the wire, $l = 1\ \text{m}$ Velocity of the wire, $v = 2\ \text{m/s}$ Magnetic field strength, $B = 0.5\ \text{T}$ Circuit resistance, $R = 6\ \Omega$ First, compute the induced motional emf ($e$): $$e = B \cdot l \cdot v = 0.5 \times 1 \times 2 = 1\ \text{V}$$ Next, calculate the mechanical power requirement using our energy conversion relationship: $$P = \frac{e^2}{R} = \frac{(1)^2}{6} = \frac{1}{6}\ \text{W}$$ Let's double check if there's a typo in the calculation or option mapping. Let's re-verify the mechanical force method: The magnetic force opposing the motion is $F = B I l$. The current is $I = \frac{e}{R} = \frac{1}{6}\ \text{A}$. $$F = B \cdot I \cdot l = 0.5 \times \frac{1}{6} \times 1 = \frac{1}{12}\ \text{N}$$ Power is Force $\times$ Velocity: $$P = F \cdot v = \frac{1}{12} \times 2 = \frac{2}{12} = \frac{1}{6}\ \text{W}$$ The mathematical derivation confirms that the rate of work done is exactly $\frac{1}{6}\ \text{W}$, which corresponds to option (B).

Step 4: Final Answer:
The rate at which work is done to keep the wire moving is $\frac{1}{6}\ \text{W}$, which corresponds to option (B).