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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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.
Step 1: Understanding the Concept:
This problem involves calculating the magnetic flux through a changing area and the motional electromotive force (EMF) induced in a conductor moving through a magnetic field.
Step 2: Key Formula or Approach:
1. Magnetic Flux (\(\Phi_B\)): The magnetic flux through a surface is defined as \(\Phi_B = B A \cos\theta\), where \(B\) is the magnetic field strength, \(A\) is the area of the surface, and \(\theta\) is the angle between the magnetic field and the normal to the surface.
2. Motional EMF (\(\mathcal{E}\)): When a conductor of length \(l\) moves with velocity \(v\) perpendicular to a uniform magnetic field \(B\), an EMF is induced across its ends, given by \(\mathcal{E} = Blv\). Alternatively, using Faraday's Law, \(\mathcal{E} = -d\Phi_B/dt\).
Step 3: Detailed Explanation: Magnetic Flux Calculation:
Let the length of the conductor PQ be \(l\) and its distance from the side RS be \(x\) at any instant.
The area of the rectangular loop PQRS is \(A = l \times x\).
The magnetic field \(\vec{B}\) is uniform and directed perpendicularly into the plane of the loop. So, the angle \(\theta\) between \(\vec{B}\) and the area vector is \(0^\circ\), and \(\cos(0^\circ) = 1\).
The magnetic flux linked with the loop is:
\[ \Phi_B = B \cdot A = B (lx) = Blx \]
Motional EMF Calculation: Method 1: Using the motional EMF formula
The conductor PQ of length \(l\) is moving with a constant velocity \(V\) perpendicular to the magnetic field \(B\).
The motional EMF induced across PQ is directly given by:
\[ \mathcal{E} = BlV \]
Method 2: Using Faraday's Law of Induction
The EMF is the rate of change of magnetic flux.
\[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(Blx) \]
Since \(B\) and \(l\) are constant, we have:
\[ \mathcal{E} = -Bl \frac{dx}{dt} \]
The conductor is moved towards the left, so the distance \(x\) is decreasing. The rate of change of position, \(dx/dt\), is the velocity. Since \(x\) decreases, \(dx/dt = -V\).
\[ \mathcal{E} = -Bl(-V) = BlV \]
Both methods give the same result for the motional EMF.
Step 4: Final Answer:
The magnetic flux is \(Blx\) and the motional EMF is \(BlV\). This matches option (C).
