The sun delivers 10⁴ W/m² of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimension (10 x 10) m²  will be

• 10⁴ W

• 10⁵ W

• 10⁶ W

• 10⁷ W

Current in a circuit falls steadily from 2.0 A to 0.0 A in 10 ms. If an average emf of 200 V is induced, then the self-inductance of the circuit is

• 1 H

• 2 H

• 1/2 H

• 2 H

The polarity of the capacitor in the situation described by Fig.

• Plate A and plate B both positive

• Plate A negative and plate B positive

• Plate A positive and plate B negative

• Plate A and plate B both negative

When a given coil is connected to a 200V, 50 Hz AC source, 1 A current flows in the circuit. The inductive reactance of the coil used is

• $50\sqrt{3} \mathrm{\Omega }$

• $200\sqrt{3} \mathrm{\Omega }$

• $150\sqrt{3} \mathrm{\Omega }$

• $100 \sqrt{3} \mathrm{\Omega }$

The magnetic flux through a circuit of resistance R changes by Δ$\mathrm{\varphi }$ in times Δt. Then, the-total amount of  electric charge that passes in any point in the circuit during this time 'Δt' will be

• $\mathrm{Q} = \frac{∆\mathrm{\varphi }}{\mathrm{R}}$

• $\mathrm{Q} = ∆\mathrm{\varphi } . \mathrm{R}$

• $\mathrm{Q} = \frac{\mathrm{R}∆\mathrm{t}}{∆\mathrm{\varphi }}$

• $\mathrm{Q} = ∆\mathrm{t} . \mathrm{R}$

The magnetic flux in a closed circuit of resistance 10 Ω varies with time t (in second) as $\mathrm{\pi }$ = (4t² − 5t + 1)Wb. The magnitude of induced current at t = 0.20 s is

• 0.12 A

• 0.38 A

• 0.34 A

• 0.12 A

The self-inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50 Hz, it should be
connected to a capacitance of

• 4µF

• 8µF

• 1µF

• 2µF

Three solenoid coils of same dimensions, same number of turns and same number of layers of windings are taken. Coil 1 with inductance L₁ was wound using a wire of resistance 11 Ω /m, coil 2 with inductance L₂ was wound using the similar wire but the direction of winding was reversed in each layer, coil 3 with inductance L₃ was wound using a superconducting wire. The self inductance of the coils L₁ , L₂ and L₃ are

• L₁ = L₂ = L₃

• L₁ = L₂ , L₃ = 0

• L₁ = L₃ , L₂ = 0

• L₁ > L₂ > L₃

89.

A current I= 5.0 A flows along a thin wire shaped as shown in given figure. The radius of the curved part of the wire is equal to R = 120 mm and the angle 2$\mathrm{\varphi }$ = 90°. Find the magnetic induction of the field at point O.

   

• 4.2 × 10^-4 T

• 8.8 × 10^-5 T

• 2.8 × 10^-4 T

• 4.2 × 10^-3 T

A small piece of metal wire is dragged across the gap between the poles of a magnet in 0.4 s. If change in magnetic flux in the wire is 8 x 10^-4 Wb, then emf induced in the wire is

• 8 × 10^-3 V

• 6 × 10^-3 V

• 4 × 10^-3 V

• 2 × 10^-3 V