Question

Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move. The conductor PQ is moved towards the left with a constant velocity V as shown in the figure. Assume that there is no loss of energy due to friction. What will be the magnetic flux linked with the loop PQRS and the motional emf?
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• A. Magnetic flux = Bl(xV); Motional emf = Blx
• B. Magnetic flux = Blx; Motional emf = (BlV)/t
• C. Magnetic flux = Blx: Motional emf = BlV
• D. Magnetic flux = B/(lx); Motional emf = BlV

Hint:

For motional EMF problems, you can often use either \(\mathcal{E} = Blv\) (if B, l, v are mutually perpendicular) or Faraday's Law \(\mathcal{E} = -d\Phi/dt\). The formula \(\mathcal{E} = Blv\) is usually quicker if applicable. Remember that the \(x\) in the flux expression represents an instantaneous position, while \(V\) in the EMF expression represents the constant velocity.

Answer 1

The Correct Option is C

Step 1: Conceptualization:
This problem addresses the calculation of magnetic flux through an evolving area and the induced electromotive force (EMF) in a conductor moving within a magnetic field.

Step 2: Governing Principles:
1. Magnetic Flux (\(\Phi_B\)): Defined as \(\Phi_B = B A \cos\theta\), where \(B\) is magnetic field strength, \(A\) is surface area, and \(\theta\) is the angle between the magnetic field and the surface normal.
2. Motional EMF (\(\mathcal{E}\)): For a conductor of length \(l\) moving with velocity \(v\) perpendicular to a uniform magnetic field \(B\), the induced EMF is \(\mathcal{E} = Blv\). Alternatively, \(\mathcal{E} = -d\Phi_B/dt\) via Faraday's Law.

Step 3: Derivation:
Magnetic Flux Determination:
Consider a conductor PQ of length \(l\) at distance \(x\) from side RS.
The rectangular loop PQRS area is \(A = l \times x\).
The magnetic field \(\vec{B}\) is uniform and perpendicular to the loop's plane, meaning \(\theta = 0^\circ\) and \(\cos(0^\circ) = 1\).
The magnetic flux through the loop is:
\[ \Phi_B = B \cdot A = B (lx) = Blx \]Motional EMF Calculation:
Method 1: Direct Motional EMF Formula
For conductor PQ of length \(l\) moving at velocity \(V\) perpendicular to magnetic field \(B\):
\[ \mathcal{E} = BlV \]Method 2: Faraday's Law of Induction
EMF equals the rate of change of magnetic flux.
\[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(Blx) \]With constant \(B\) and \(l\):
\[ \mathcal{E} = -Bl \frac{dx}{dt} \]The conductor moves leftward, decreasing \(x\). Thus, \(dx/dt = -V\).
\[ \mathcal{E} = -Bl(-V) = BlV \]Both methods yield the same motional EMF.

Step 4: Conclusion:
The magnetic flux is \(Blx\), and the motional EMF is \(BlV\), consistent with option (C).