Question

Two straight conducting plates form an angle θ where their ends are joined. A conducting bar in contact with the plates and forming an isosceles triangle with them, starts at the vertex at time t = 0 and moves with constant velocity v to the right as shown in the figure. A magnetic field B points out of the page. The magnitude of emf induced at t = 1 second will be:

• A. \( Bv \tan \frac{\theta}{2} \)
• B. \( 2Bv^2 \tan \frac{\theta}{2} \)
• C. \( 2Bv^2 \cot \frac{\theta}{2} \)
• D. \( 2Bv^2 \sin \frac{\theta}{2} \)

Hint:

The induced emf in a moving conductor depends on the velocity of the conductor and the angle between the conducting plates, as wellas the magnetic field strength.

Answer 1

The Correct Option is B

The motion of the conducting bar creates a change in the area enclosed by the conducting plates, which results in an induced emf. The emf (\(\mathcal{E}\)) is given by Faraday’s law of electromagnetic induction: \[ \mathcal{E} = \frac{d\Phi}{dt}, \] where \(\Phi\) is the magnetic flux. The magnetic flux is: \[ \Phi = B \cdot \text{Area}. \] At \(t = 1\) second, the distance moved by the bar is: \[ x = vt, \] where \(v\) is the velocity of the bar and \(t = 1\). Step 1: Area of the Triangle Formed The height of the triangle at time \(t\) is: \[ h = x \tan \frac{\theta}{2} = vt \tan \frac{\theta}{2}. \] The area of the triangle is: \[ \text{Area} = \frac{1}{2} \cdot \text{Base} \cdot \text{Height}. \] The base of the triangle is \(2x\), so: \[ \text{Area} = \frac{1}{2} \cdot 2x \cdot h = x^2 \tan \frac{\theta}{2}. \] Step 2: Magnetic Flux The magnetic flux through the triangle is: \[ \Phi = B \cdot \text{Area} = B \cdot x^2 \tan \frac{\theta}{2}. \] Step 3: Induced emf The induced emf is the rate of change of flux: \[ \mathcal{E} = \frac{d\Phi}{dt} = \frac{d}{dt} \left( B \cdot x^2 \tan \frac{\theta}{2} \right). \] Since \(\tan \frac{\theta}{2}\) and \(B\) are constants, we differentiate \(x^2\) with respect to \(t\): \[ \mathcal{E} = B \cdot \tan \frac{\theta}{2} \cdot \frac{d(x^2)}{dt}. \] \[ \frac{d(x^2)}{dt} = 2x \cdot \frac{dx}{dt} = 2xv. \] Substitute \(x = vt\) and \(t = 1\): \[ \mathcal{E} = B \cdot \tan \frac{\theta}{2} \cdot 2(vt)v. \] At \(t = 1\): \[ \mathcal{E} = 2Bv^2 \tan \frac{\theta}{2}. \] Conclusion The induced emf at \(t = 1\) second is: \[ \boxed{2Bv^2 \tan \frac{\theta}{2}}. \]