In the inductive circuit,

rms value of current, I = 15.92 A

rms value of voltage, V = 220 V

Hence, the net power absorbed can be obtained by the relation,

P = VI cos φ

Where, φ = Phase difference between V and I.

For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., φ= 90°.

Hence, P = 0 i.e., the net power is zero.

In the capacitive circuit, rms value of current, I = 2.49 A, rms value of voltage, V = 110 V Hence, the net power absorbed can be obtained as:

P = VI Cos φ

For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e.,

φ = 90°.

Hence, P = 0 i.e., the net power is zero.

Inductance of inductor, L = 44 mH = 44 x 10⁻³ H

Supply voltage, V = 220 V,

Frequency, ν = 50 Hz,

Angular frequency, ω = 2πν

Inductive reactance, X_L= ωL = 2πνL = 2π x 50 x 44 x 10^-3 Ω

rms value of current is given as:

I = V/X_L =220/(2π x 50 x 44 x 10^-3) =15.92 A

Hence ,the rms value of current in the circuit is 15.92 A

Take a small element dy in the loop at a distance y from the long straight wire [as shown in the given figure).

Magnetic flux associated with element dy, dφ = BdA

Where, dA = Area of element dy = a dy

B = Magnetic field at distance y

=μ0I/2πy

I = Current in the wire

μ0= Permeability of free space = 4π x 10⁻⁷

Therefore ,dφ = (μ0Ia/2π) x (dy/y)

φ = μ0Ia/2π ⌠ dy/y                                              _{{ NOTE- “⌠” is sign of integration}}

v tends from x to a + .x.

Therefore ,φ = μ0Ia/2π _x⌠^a+x dy/y

=μ0Ia/2π  [log_e y]_x^a+x

=μ0Ia/2π  [log_e(a+x)/x]

For mutual inductance M. the flux is given as: φ = MI

Therefore ,MI = μ0Ia/2π  [log_e(a/x+1)]

M=μ0a/2π  [log_e(a/x+1)]

Ans (b).

Emf induced in the loop, e = B’av

= (μ0I/2π)x av

Given, I = 50 A

x = 0.2 m

a = 0.1 m

v= 10 m/s

e = (4π x 10⁻⁷ x 50 x 0.1 x 10 )/(2π x 0.2)

e = 5 x 10⁻⁵ V

Length of the rod, l = 15 cm = 0.15 m

Magnetic field strength, B = 0.50 T

Resistance of the closed loop, R = 9 mΩ = 9 x 10⁻³ Ω

Ans (a).

Induced emf = 9 mV; polarity of the induced emf is such that end P shows positive while end Q shows negative ends. Speed of the rod, v = 12 cm/s = 0.12 m/s Induced emf is given as:

e = Bvl

= 0.5 x 0.12 x 0.15 = 9 x 10-³ v = 9 mV

The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

Ans (b).

Yes; when key K is closed, excess charge is maintained by the continuous flow of current.

When key K is open, there is excess charge built up at both ends of the rods.

When key K is closed, excess charge is maintained by the continuous flow of current.

Ans (c).

Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.

There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

Ans (d).

Retarding force exerted on the rod, F = IBl Where,

I = Current flowing through the rod

= e/R = (9 x 10-³ )/(9 x 10-³ ) =1 A

Therefore, F = 1 x 0.5 x 0.15 =75 x 10-³ N

Ans (e).

9 mW; no power is expended when key K is open.

Speed of the rod, v = 12 cm/s = 0.12 m/s

Hence, power is given as:

P = Fv

= 75 x 10-³ x0.12 = 9×10-³ W = 9 mW

When key K is open, no power is expended.

Ans (f).

9 mW; power is provided by an external agent.

Power dissipated as heat = I² R = (l)² x 9x 10-³ = 9 mW.

The source of this power is an external agent.

Ans (g).

Zero

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.