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**Q 1. ****A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant b, the correct equivalence would be** **[2020]**

** **

** Ans:** (4)**Solution:**

Given LCR circuit

We know that equation of motion of damped oscillation is

(1)

Now in the LCR circuit, voltage drop along the circuit is

We know that

So

..(2)

Comparing Eqs. (1) and (2), we get

m = L , b= R, K = 1/C**Q 2. ****An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 Ω resistor. The ratio of the currents at time t =∞ and t = 40 s is close to (Take e ^{2} = 7.389)(1) 1.06(2) 1.15(3) 1.46(4) 0.84**

Solution:

Given that, L = 10 mH = 10 × 10

We know that, current at instant time t is given by

where

I

And

So,

I = 4[1 - exp( - 500t )]

at t =∞, we have

I_{∞} = 4[1 - exp ( -∞] = 4(1-0) = 4

at t =40, we have

I_{40 }= 4[1 - exp (-500 x 400)] = 4[1-exp (-20000)]

= 4[1-(e^{2)-10000}] = 4[1 - (7.389)^{-10000}]

So,

Therefore I_{∞}/I_{40} is slightly greater than 1.**Q 3. As shown in the figure, a battery of emf ε is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = t _{c} (t_{c} is the time constant of the circuit) is [2020]** (1)

Ans:

Charge flow for time t

Here t

So,

Current flow in the circuit at any instant t is given by

Here I

So,

(1) 0.52cos(314t)

(2) 10cos(314t)

(3) 5.2cos(314t)

(4) 0.52sin(314t)

Ans:

Given that LC series circuit having

L = 40mH = 40 x 10

C = 100μF = 100 x 10

V(t) = 10sin (314t)

So,

ω = 314

X

Z = X

Since, X

Now, maximum current is given by

Therefore, current in the circuit is given by

Ans:

Given that

Standard equation of magnetic field of EM wave is

So,

B

we know that

E

= 3 x 10

Standard equation of electric field of EM wave is

Therefore,

(1) Parallel to

(2) zero

(3) antiparallel to

(4) parallel to

Ans:

Given that,

at t = 0, position of charged particle (x,y,z)

Instantaneous velocity,

We know force is given by

at t = 0,

We know that speed of electromagnetic waves is given by

So, at t = 0,

Therefore,

So, force is antiparallel to

given that,

We know that

At t = 0, a particle of charge q is at origin with a velocity , (c is the speed of light in vacuum). The instantaneous force experienced by the particle is [2020]

Ans:

Given that

For given electric fields of electromagnetic waves, there corresponding magnetic fields are

Force on a charge particle in these two electromagnetic waves is given by

Ans:

Direction of propagation of EM wave is

Direction of polarization of electric filed is

So, direction of polarization of magnetic field is

Therefore, direction of polarization of magnetic field is

(1) Direction of I

(2) At t = 0.25 s direction of I

(3) Direction of I

(4) At t = 0.5 s direction of I

Ans:

Given that

I(t) = I

Induced emf is given by

where ϕ is flux

Solenoid magnetic flux is given by

So,

For t = 0.5 s, V

So emf will be zero at t = 0.5 s

At t = 0.5 s, direction of I

(1) 2.5 s and 7.5 s

(2) 2.5 s and 5.0 s

(3) 5.0 s and 7.5 s

(4) 5.0 s and 10.0 s

Ans:

We know, magnetic flux is given by

Where B is magnetic field and A is area

And emf induced is given by

emf will be maximum when

and emf will be minimum when ωt = π, 2π, ....

So, magnitude of induced emf will be maximum and minimum at 2.5 s and 5.0 s respectively.

[2020]

(1) 56 μV

(2) 28 μV

(3) 48 μV

(4) 36 μV

Ans:

Given that

B

B

t= 5s

Induced EMF is given by

Ans:

Given that

V =100V, I

Voltage across inductor is given by

Ans:

Magnetic moment of circular loop carrying current I and radius a is given by

Where is the area vector of the circular loop.

Given conditions is shown in the following figure

Torque on a current carrying circular loop in a magnetic field is given by

And

Where angular acceleration of the circular loop.

So, Iα = MBsin θ

For small θ , we have

We know that, for SHM

α =-ω^{2}θ

So,

We know that

T = 2π/ω**Q ****15. A conducting circular loop made of a thin wire, has area 3.5 × 10 ^{−3} m^{2} and resistance 10 Ω. It is placed perpendicular to a time dependent magnetic field B(t) = (0.4 T) sin (50πt). The field is uniform in space. Then, the net charge flowing through the loop during t = 0 s and t = 10 ms is close to [2019]** (1)

(1) 14 mC

(2) 7 mC

(3) 21 mC

(4) 6 mC

Ans:

We have, A = 3.5 × 10

=(1/10) x 3.5 x 10

= 1.4 × 10

(1) 12 mV

(2) 6 mV

(3) 1 mV

(4) 2 mV

Ans:

Potential difference between two faces = E × l

= (v × B)l

= 6 × 0.1 × 2 × 10

= 12 mV

Ans:

Self-Inductance is

L = μ

Mutual inductance is

M = μ

(1) decreases by a factor of 9.

(2) increases by a factor of 27.

(3) increases by a factor of 3.

(4) decreases by a factor of 9√3

Ans:

The self-inductance of the coil is

⇒ N/l = constant

⇒ N ∝ l

⇒ L ∝ lA

⇒ L ∝ la

Therefore, self-inductance will increase by a factor of 3.

(1) 1.5 × 10

(2) 1.1 × 10

(3) 2.5 × 10

(4) 0.3 × 10

Ans:

Induced emf = Bvl sin θ

Since, θ = 45°

(1) 50 A

(2) 45 A

(3) 35 A

(4) 25 A

Ans:

(1) 5.65 × 10

Ans:

Given R = 60 Ω,

Inductive Reactance

X

= 2π

= 2 × π × 50 × 20 × 10

= 6.28 Ω

Now,

X

= 20.24

X

Impedance

Power consumed by the AC

Hence, energy dissipated in the circuit in 60s = 5.17 × 10

(1) 740 J

(2) 437.5 J

(3) 540 J

(4) 637.5 J

Ans:

Induced emf,

Therefore, change in energy of inductance is

(1)

(2)

(3)

(4)

*Disputed question – As more than one option is correct [options (1), (2), (4) are correct

We have,

Now, integrating both the sides, we get

during decay

[2019]

(1) 5.5 A

(2) 7.5 A

(3) 3 A

(4) 6 A

Ans:

Let I be the current in the circuit.

Ideal inductor will behave like zero resistance long time after switch is closed.

I = 2E/R

⇒ I = (2 x 15) / 5

⇒ I = 6A

[2019]

**(1) 60°(2) 30°(3) 90°(4) 0°Ans:** (3)

Given V = 200√2 sin100t

ω = 100

X

Thus, θ

⇒θ

So, the phase difference = 90° + 60° = 150°

If R

(1) 18.9 x 10

(2) 2.1 x 10

(3) 6.3 x 10

(4) 18.9 x 10

Ans:

Let be the magnetic field, at a particular point

So,

where c is the speed of light.

Unit vector

(1) U

(2) U

(3) U

(4) U

Ans:

The average electric energy density is

The average magnetic energy density is

In electromagnetic wave, the electric and magnetic field vary sinusoidally in free space so, in above expression E and B are replaced by their rms values.

Therefore,

Since energy density of electric and magnetic field is equal.

Hence, U_{B} = U_{E}**Q 28. If the magnetic field of a plane electromagnetic wave is given by ( the speed of light = 3 × 10 ^{8 }m/s)then the maximum electric field associated with it is [2019](1) 6 × 10^{4} N/C(2) 3 × 10^{4} N/C(3) 4 × 10^{4} N/C(4) 4.5 × 10^{4} N/CAns:** (2)

speed of light is

c = 3 × 10

Magnetic field is

Maximum magnetic field, B

Therefore, electric field is

E

= 3 × 10

= 3 × 10

(1)

(2)

(3)

(4)

Ans:

Electric field is

since, is direction of waves

So, direction of will be along

Let and unit vector in direction of propagation of EM wave is

Therefore,

Ans:

As we know that,

For transparent medium

From Eq. (1) and Eq. (2), we have

We know that, intensity in term of electric field is

From Eq. (3) and Eq. (4), we have

Similarly, in terms of magnetic field

From Eq. (5) and Eq. (6), we get

(Given: Permittivity of space ε

(1) 2 kV/m

(2) 0.7 kV/m

(3) 1 kV/m

(4) 1.4 kV/m

Ans:

Intensity of electromagnetic wave is given by

E = √2 x 10

E =1.4 kV/m

(1) 50, 10

(2)

**(3) (4) 50, 0Ans: **(2)

(1)

**(2) **

**(3) **

**(4) Ans:** (1)

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