Question

Figure shows a rectangular frame situated in a constant magnetic field. A wire BC of length \(1\text{ m}\) is moved out with velocity \(4\text{ m/s}\). Magnetic field strength is \(0.15\text{ T}\). Force acting on the wire \(BC\) is

• A. \(18\text{ N}\)
• B. \(1.8\text{ N}\)
• C. \(0.18\text{ N}\)
• D. \(0.018\text{ N}\)

Hint:

Moving conductor shortcut: $F = \frac{B^2 \ell^2 v}{R}$

Answer 1

The Correct Option is D

Concept:
• Induced emf: \(\varepsilon = B \ell v\)
• Current: \(I = \frac{\varepsilon}{R}\)
• Magnetic force: \(F = B I \ell\)

Step 1: Calculate induced emf. \[ \varepsilon = B \ell v = 0.15 \times 1 \times 4 = 0.6 \text{ V} \]

Step 2: Find current. \[ I = \frac{\varepsilon}{R} = \frac{0.6}{5} = 0.12 \text{ A} \]

Step 3: Force on wire. \[ F = B I \ell = 0.15 \times 0.12 \times 1 = 0.018 \text{ N} \]

Step 4: Conclusion.
Force on wire = \(0.018 \text{ N}\) Final Answer: Option (D)