Physics M.C.Q [1M]

1. $\frac{\text{BL}\omega\text{dt}}{\rho}$

Explanation:

$\text{emf},\in=\frac{\text{d}\phi}{\text{dt}}=\text{BL}^2\omega$

$\text{I}=\frac{\in}{\text{R}}=\frac{\text{BL}^2\omega}{\rho\frac{\text{L}}{\text{dt}}}=\frac{\text{BL}\omega\text{dt}}{\rho}$

4. $\text{Zero.}$

Explanation:

Induced emf is AB is Bvl and Induced emf is DC is also Bvl.

Net emf in the closed circuit (loop) is zero.

So induced current in the loop is zero.

3. $\frac{4\pi}{3}\text{volt}$

Explanation:

$\text{Max.emf = nAB}\omega$

$=50\times0.008\times0.05\times2000\frac{2\pi}{60}=\frac{4\pi}{3}\text{V}$

2. the current

Explanation:

Self-inductance of a solenoid is given by $\text{L}=\frac{\mu_0\text{N}^2\text{A}}{1}$

where, N = number of turns in solenoid

l = length of the solenoid

A = area of one turn of solenoid.

The self-inductance of solenoid does not depend on current passing through it.

3. 15KV

Explanation:

Flux linking the secondary coil due to current in primary = Mi₁​ 

induced emf in the secondary coils = rate of change of flux

$=\text{M}\frac{\triangle\text{i}_1}{\triangle\text{t}}$

$=5\times\frac{3-0}{0.001}$

$=15\text{kv}$

2. The potential difference between C and the rim is $\frac{1}{3}\text{Br}^2\omega$
3. C is at a higher potential than the rim.

1. +12V

Explanation:

induced emf = Blv = 12V. It is induced in the northward direction by right hand rule $\big(\text{emf}=\overrightarrow{\text{V}}\times\overrightarrow{\text{B}}\big)$

therefore if south end of pole has potential of 0V, north end will have a potential of 12V

3. 0.06 V

Explanation:

emf = rate of change of flux × number of turns $=\frac{12\times10^3\times10^{-8}}{0.2}\times100=0.06\text{V}$

emf = rate of change of flux × number of turns.

2. 0.4H

Explanation:

$\text{emf}=\text{L}\frac{\text{di}}{\text{dt}}$

$\text{emf}=\text{L}\frac{\triangle\text{i}}{\triangle\text{i}}$

So, $\text{L}=\text{emf}\times\frac{\triangle\text{t}}{\triangle\text{i}}$

$=32\times\frac{0.1}{8}$

$=0.4\text{H}$

2. BLv

Explanation:

Force on a charge q in the rod = qvB

electric field inside the rod due to displacements of charges due to force by relative motion in magnetic field $=\frac{\text{emf}}{1}$

As the rod moves in constant velocity, net force of constituent charge q in the rod = 0 

Therefore,

$\text{q}\frac{\text{emf}}{1}=\text{qvB}$

$\Rightarrow\text{emf}=\text{BIv}$

4. 4 × 10^-4C 

Explanation:

Initial flux through the coil, $=\text{BA}\cos\theta$

$=\text{BA}\cos0^\circ$ as the coil is perpendicular to magnetic field

= BA Wb

Final flux after the rotation,

$=\text{BA}\cos180^\circ$ the coil is rotated through 180^°

=-BA Wb

Therefore, estimated value of the induced emf (E) as per Faraday's law is,

$\text{E}=\frac{\text{N}}{\text{t}}=\frac{\text{N}(\text{BA-(-BA}))}{\text{t}}=\frac{\text{2NBA}}{\text{t}}........(\text{i})$

Further, $\text{E}=\text{IR}=\frac{\text{QR}}{\text{t}}............(\text{ii})$

From (i) & (ii)

$\text{Q}=\frac{\text{2NBA}}{\text{R}}=\frac{2\times2000\times4\times10^{-5}\times500\times10^{-4}}{20}=4\times10^{-4}\text{C}$

4. 0.1H

Explanation:

lf e is the induced e.m.f. in the coil, then $\text{e}=\text{-L}\frac{\text{di}}{\text{dt}}$

$\therefore\text{L}=-\frac{\text{e}}{\frac{\text{di}}{\text{dt}}}$

Substituting values, we get $\text{L}=\frac{-8\times0.05}{-4}=0.1\text{H}$

2. l decreases and A increases.

Solution:

Key concept: The self inductance L of a solenoid depends on various factor like geometry and magnetic permeability of the core material.

$\text{L}=\mu_\text{r}\mu_0\text{n}^2\text{Al}$

where, $\text{n}=\frac{\text{N}}{\text{l}}$ (no. of turns per unit length)

1. No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2. Area of cross section: Larger the area of cross section of the solenoid, larger is its self inductance.
3. Permeability of the core material. The self inductance of a solenoid increases μr times if it is wound over an iron core of relative permeability μr.

The long solenoid of cross-sectional area A and length l, having A turns, filled inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) then its self inductance is $\text{L}=\mu_\text{r}\mu_0\text{n}^2\frac{\text{A}}{\text{l}}$.

So, the self inductance L of a solenoid increases as l decreases and A increases because L is directly proportional to area and inversely proportional to length.

Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/permeability of the medium.

3. 4 mH

Explanation:

Given,

L¹​ = 2mH

L² ​= 8mH

The mutual inductance between coil is

$\text{M}=\sqrt{\text{L}_1\text{L}_2}$

$\text{M}=\sqrt{2\times8}=16\text{mH}$

$\text{M}=4\text{mH}$

2. 18m²

Explanation:

Flux through a circular coil $\phi=\text{NBA}\cos\omega\text{t}$

Voltage required $\in=\frac{-\text{d}\phi}{\text{dt}}$

$\Rightarrow9=\text{NBA}\omega\sin\omega\text{t}$

$9=\frac{8\times10^{-5}\times\text{A}\times30\times2\pi\times2000}{60}$

$\text{A}=\frac{9\times10^5}{50\times10^3}=18\text{m}^2$

3. high inductance

Explanation:

On removal of load from the circuit, the circuit suddenly becomes an open circuit.

Thus $\frac{\text{di}}{\text{dt}}\rightarrow\infty$

For sparking, high voltage must appear across the open ends. This will happen only in case of an inductor as the voltage drop across the

inductor is $\text{L}\frac{\text{di}}{\text{dt}}$

Therefore, the circuit has high inductance.

2. $\frac{\text{mg}}{\text{il}}\tan\theta$

Explanation:

The magnetic force due to the conducting rod of length l and mass m is,

F_m​ = i(l × B), the angle between l and B is 90°.

F_m ​= ilB

The horizontal component of F_m​ is

$\text{F}=\text{F}_\text{m}\cos\theta$

The force due to gravity is F = mg and its horizontal component is

$\text{F}''=\text{mg}\sin\theta$

From the above figure,

$\text{mg}\sin\theta=\text{ilB}\cos\theta$

$\text{B}=\frac{\text{mg}}{\text{il}}\tan\theta$

3. $\frac{80\text{A}}{\text{s}}$

Explanation:

$\text{emf}=\text{L}\frac{\text{di}}{\text{dt}}$

$\frac{\text{di}}{\text{dt}}=\frac{\text{emf}}{\text{L}}$

$=\frac{8}{0.1}$

$=\frac{80\text{A}}{\text{s}}$

2. 40H

Explanation:

Coefficient of mutual induction will be the ratio of the flux linked with the secondary coil and the current primary coil.

$\text{M}=\frac{200}{5}=40\text{H}$

3. Move away from the solenoid.

Explanation:

$\text{e}=-\text{L}\frac{\text{di}}{\text{dt}}$

Current flow in the CKt is clock wise direction, due to Mutual Induction current flow in the loop anti clockwise direction. The net force applied on the loop in east direction. So we can say that the ring will move away from the solenoid.

1. $\text{zero}$

Explanation:

Emf at both end is same $=\frac{1}{8}\text{Bwl}^2$

So the potential difference between the two ends of therod is zero.

1. $\text{L}=\Big(-\frac{\text{dI}}{\text{dt}}\Big)$

Explanation:

Inductance is the property of a conductor by which a change in current flowing through it induces (creates) a voltage (electromotive force) in both the conductor itself (self-inductance) and in any nearby conductors (mutual inductance). By Lenz's law the induced voltage opposed the the change in current. Hence inductance is defined as.

$\text{L}=\Big(-\frac{\text{dI}}{\text{dt}}\Big)$

4. 0.2 A

Explanation:

$\text{emf}=-\frac{\text{d}\phi}{\text{dt}}=-12\text{t}+5$

$\text{Current}=-\frac{\text{emf}}{10}=-1.2\text{t}+0.5$

$\text{Current at t}=0.25\text{s},$

$=1.2\times0.25+0.5=0.2\text{A}$