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: \[ \boxed{2Bv^2 \tan \frac{\theta}{2}}. \]
Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \).
A circular coil of diameter 15 mm having 300 turns is placed in a magnetic field of 30 mT such that the plane of the coil is perpendicular to the direction of the magnetic field. The magnetic field is reduced uniformly to zero in 20 ms and again increased uniformly to 30 mT in 40 ms. If the EMFs induced in the two time intervals are \( e_1 \) and \( e_2 \) respectively, then the value of \( e_1 / e_2 \) is:
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: