Magnetism And Matter - Class 12 Engineering Physics - Extra Questions

The magnetic moment of the loop is given by 

In the given figure the current flow clockwise in left of the coil  but instantly when key inserted that flux passes through the coil then that moment coil (right) oppose the flux and induced current in the right coil,therefore galvanometer deflect left after that current in coil (left) is zero and Hence galvanometer shows zero deflection

Hint: Lenz law is based on Faraday's law of Induction.

Lenz's law states that when an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such, that it produces an current that's magnetic field opposes the change which produces it.
The negative sign used in Faradays law of electromagnetic induction, indicates that the induced emf $$\left( \varepsilon  \right)$$ and the change in magnetic flux $$\left( \delta \phi B \right)$$ have opposite signs.
$$\varepsilon =-N\frac { \delta \phi B }{ \delta t }$$
Where,
$$\left( \varepsilon  \right)$$  = Induced emf
$$\left( \delta \phi B \right)$$ = change in magnetic flux
N = No of turns in coil
Hence, the e.m.f. induced in the coil becomes more when the numbers of turns in the coil are made large and the increased induced e.m.f. opposes the cause with a greater force which produces it.

$$\bigodot$$ is increasing. Hence $$\bigotimes$$ is produced by the induced current. So, it is clockwise.

As you know according to Faraday's induction law or Lenz's law, an induced current creates a magnetic field that opposes any change in the magnetic flux of the ring. So your answer is that the current in the ring flows in the same direction as the current in the loop connected to the source. When the current increases it creates a magnetic flux which goes through the other loop too. So a current should be induced in that loop which opposes this magnetic flux. Try to imagine the loops side by side, if the magnetic field created by loop A is coming out of the page when it enters loop B it is going into the page. So if the current in loop B is to oppose this magnetic field, it should create a magnetic field which would come out of the page so it should be in the same direction as the current in loop A. 

$$\bigotimes$$ magnetic field passing through loop is increasing. Hence induced current will produce magnetic field. So, induced current should be anti-clockwise.

By lenz's law, induced current flows in a direction to oppose the cause.

The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:

(a) The direction of the induced current is along qrpq.

(b) The direction of the induced current is along prqp.

(c) The direction of the induced current is along yzxy.

(d) The direction of the induced current is along zyxz.

(e) The direction of the induced current is along xryx.

(f) No current is induced since the field lines are lying in the plane of the closed loop.

According to the Lenz's law, the current induced will oppose the change in flux through the loop.

Lenz's law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux and to exert a mechanical force opposing the motion.

Lenz's law states that when an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such, that it produces an current that's magnetic field opposes the change which produces it.
The negative sign used in Faraday's law of electromagnetic induction, indicates that the induced emf ( $$\epsilon$$ ) and the change in magnetic flux ( $$\delta \phi_B$$ ) have opposite signs.
$$\epsilon- N\frac{d\phi_B}{dt}$$
B is the magnetic field, N is number of turns in coil.
Lenz's law obeys the law of conservation of energy and if the direction of the magnetic field that creates the current and the magnetic field of the current in a conductor are in same , then these two magnetic fields would add up and produce the current of twice the magnitude  creating more magnetic field. This will  cause more current and leads to violation of the law of conservation of energy.

If N-pole of a magnet is move toward the coil, an induce current produced in the coil and the direction of induced current will be such that it opposes the reaction of N-pole of magnet towards the coil. Hence N pole is formed at the face of the coil, which is towards the magnet. Therefore direction of induced current will be anticlockwise from the magnet.
If N-pole of magnet is more away from the coil; an in danced current produced in the coil and the direction of induced current will be such that it opposes the motion of N-pole away from the coil. Hence S-pole is formed at the face of the coil, which is towards the magnet. Therefore direction of induced current will be current of manual magnet.

The induced current has no direction of its own always flows in a direction so that it opposes the cause.For example, if a magnet is dropped through a copper ring, the induced current flows in the ring in the clockwise direction, till the magnet does not come out of the ring from the other side.But as soon as the magnet leaves the ring the direction of the current will become anti-clockwise in according to Lenz's law.it shows that an induced current has no direction of its own.

Lenz's law states that when an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such, that it produces a current whose magnetic field opposes the change which produces it.

Mathematically,

( : Emf ; N : Number of turns ; : Magnetic Flux)
Here the negative sign is due to the Lenz's law.

Lenz's law gives the magnitude and direction of the electric current induced in the coil due to change in magnetic flux though that coil.

Lenz's law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux and to exert a mechanical force opposing the motion. 

As per Kinetic energy-work theorem(Work energy principle)

The formula of magnetic moment is $$M =NIA$$  

Two electrons move in parallel direction, at distance $$r$$ apart with speed $$=v$$ Let it move 'dx', So,

for a revolviing electron $$.$$

Combining Eq. 32-27 with Eqs. 32-22 and 32-23, we see that the energy difference is
                                  $$ΔU =2μ_BB$$  
where $$μ_B$$ is the Bohr magneton (given in Eq. 32-25). With $$ΔU = 6.00 \times 10^{−25}$$ J, we
obtain $$B = 32.3\, mT$$ .

(a) It is known  $$E=\dfrac{e}{4\pi \epsilon_0 r^2}$$
                      $$E=\dfrac{e}{4\pi \epsilon_0 r^2}=\dfrac{(1.60 \times 10^{-19}\,C)(8.99 \times 10^9\,N.m^2/C^2)}{(5.2 \times 10^{-11}\,m)^2}=5.3 \times 10^{11}\,N/C$$  

(b) It is known : $$B=\dfrac{\mu_0}{2\pi}\,\dfrac{\mu_p}{r^3}$$

As we know,

(a) It is known , $$μ_{orb,z}= –mμB$$

Lenz's law: According to this law " the direction of induced current in a closed circuit is always such as to oppose the cause that produces it". 
The direction of induced current in a circuit is such that it opposes the very cause which generates it. Yes, an emf will be induced at its ends. Justification: As the metallic rod falls down, the magnetic flux due to vertical component of Earth's magnetic field keeps on changing.

In a superconductor there is no resistance, Hence,
$$L \dfrac {dI}{dt} = + \dfrac {d\Phi}{dt}$$
So integrating, $$I = \dfrac {\triangle \Phi}{L} = \dfrac {\pi a^{2} B}{L}$$
because, $$\triangle \Phi = \Phi_{f} - \Phi_{i}, \Phi_{f} = \pi a^{2}B, \Phi_{i} = 0$$
Also, the work done is, $$A = \int \xi I\ dt = \int I\ dt \dfrac {d\Phi}{dt} = \dfrac {1}{2} L\ I^{2} = \dfrac {1}{2} \dfrac {\pi^{2} a^{4} B^{2}}{L}$$ .

In ground state $$(n=1)$$ according to Bohr's theory:
$$mvR=\dfrac {h}{2\pi}$$ or $$v=\dfrac {h}{2\pi mR}$$

The potential energy of circular loop in the magnetic field is $$-MB cos\theta$$ , where $$\theta$$ is angle between normal to plane of coil and B. Initially given $$\theta=0^o$$
$$\therefore$$ Initial potential energy,  $$U_i=NiAB \ cos \ 0^o=-NiAB$$
When coil is turned through $$180^o; \theta=180^o$$ , therefore, final potential energy
$$U_f=-NiAB cos 180^o=NiAB$$
$$\therefore$$ Required work, W $$=$$ gain in potential energy
                                  $$=U_f-U_i\\ =NiAB-(-NiAB)\\ =2NiAB$$

emf in the loop BCDH = $$m^2 \times 10 V$$ . As B increases with time, to counter this effect, by Lenz's law, current in this loop flows in the direction CBHDC(counterclockwise).value of curren in this loop = $$i_1=\frac{10m^2}{m^{-1}4m} = \frac{10m^2}{4} A$$

emf in the loop  DEFH = $$m^2 \times 10 V$$ . As B increases with time, to counter this effect, by Lenz's law, current in this loop flows in the direction DHFED(counterclockwise). Value of current in this loop = $$i_2=\frac{10m^2}{m{-1}4m} =\frac{10m^2}{4} A$$

emf in the loop ABFG = $$2m^2 \times 10 V$$ .As B increases with time, to counter this effect, by Lenz's law current in this loop flows in the direction BAGFB(counterclockwise). Value of current in this loop = $$i_3=\frac{20m^2}{m^{-1}6m} = \frac{20m^2}{6}$$

Therefore net current in each segment is as in figure.

Heat is produced at the rate = $$i_3^2 R_{AB}+ i_1^2 R_{BC} + i_1^2 R_{CD}+ i_2^2 R_{DE} + i_2^2 R_{EF} + i_3^2 R_{FG}+ i_3^2 R_{AG} + (i_1-i_2)^2 R_{DH} + (i_1-i_3)^2 R_{BH} + (i_2-i_3)^2 R_{FH}=\frac{425 m^4}{6} W$$

Induced emfs are 10V, 10V and 20 V as shown.
Top left loop: $$10=(I_1-I_2) \times 1+ (I-I_2) \times 2- I_2$$

$$\Rightarrow 10=2l-4I_2+I_1$$ ....(i)

Top right loop: $$10=(I-I_1) \times 1+ (I-I_1) \times 1-(I_1-I_2)\times 1 -I_1\times 1$$

$$\Rightarrow 10=2l-4I_2+I_1$$ ....(ii)

from (i) and (ii) $$I_1=I_2$$

Bottom loop: $$20=2I+1I+1I_1+1I_2+1I$$

$$\Rightarrow 20= 4I_1+I_2$$

$$\Rightarrow 20= 4I+2I_1$$ ....(iii)

from (ii) $$10=2I-3I_1 \Rightarrow  20=4I-6I_1$$ ...(iv)

From (ii) and (iv) $$I_1=0, so I_2=0$$ then from (iii) $$I=5a$$ so current of 5A flows in outer branch and no curent in the inner branches.

Heat produced = $$I^2 R_{outer branches}=5^2 \times 8=200W$$   

The magnetic dipole moment is $$\vec{\mu} = \mu (0.60 \hat{i}-0.80 \hat{j})$$ , where

$$\mu = NiA$$

$$= Ni \pi r^{2}$$

$$=1(0.20A)(\pi)(0.080m)^{2}$$

$$= 4.02 \times 10^{-4}A \cdot m^{2}$$ .

Here i is the current in the loop, $$N$$ is the number of turns, $$A$$ is the area of the loop, and $$r$$ is its radius.

The orientation energy of the dipole is given by

$$U = -\vec{\mu} \cdot \vec{B}$$

$$=-\mu(0.60\hat{i}-0.80\hat{j}) \cdot (0.25\hat{i}+0.30\hat{k})$$

$$= -\mu (0.60)(0.25)$$

$$=-0.15\mu$$

$$=-6.0 \times 10^{-4}J$$ .

Here $$\hat{i} \cdot \hat{i} =1$$ , $$\hat{i} \cdot \hat{k} =0$$ , $$\hat{j} \cdot \hat{i} =0$$ and $$\hat{j} \cdot \hat{k} =0$$ are used.

We note that 1 gauss $$=10^{-4}$$ T.
The amount of charge is

$$q(t) =\dfrac{N}{R}\left[B A \cos 20^{\circ}-\left(-B A \cos 20^{\circ}\right)\right]=\dfrac{2 N B A \cos 20^{\circ}}{R}$$

$$=\dfrac{2(1000)\left(0.590 \times 10^{-4} \mathrm{T}\right) \pi(0.100 \mathrm{m})^{2}\left(\cos 20^{\circ}\right)}{85.0 \Omega+140 \Omega}=1.55 \times 10^{-5} \mathrm{C}$$

Note that the axis of the coil is at
$$20^{\circ},$$ not $$70^{\circ},$$ from the magnetic field of the Earth.

From the datum at $$t=0$$ in Fig. b
we see $$0.0015 \mathrm{A}=V_{\text {battery }} / R,$$ which implies that the
resistance is

$$R=(6.00 \mu \mathrm{V}) /(0.0015 \mathrm{A})=0.0040 \Omega$$

Now, the value of the current during
$$10 \mathrm{s}<t<20 \mathrm{s}$$ leads us to equate

$$\left(V_{\text {battery }}+\varepsilon_{\text {induced }}\right) / R=0.00050 \mathrm{A}$$

This shows that the induced emf is $$\varepsilon_{\text {induced }}=-4.0 \mu \mathrm{V}$$ . Now we use Faraday's law:

$$\varepsilon=-\dfrac{d \Phi_{B}}{d t}=-A \dfrac{d B}{d t}=-A a$$

Plugging in $$\varepsilon=-4.0 \times 10^{-6} \mathrm{V}$$ and $$A=5.0 \times 10^{-4} \mathrm{m}^{2},$$ we
obtain $$a=0.0080 \mathrm{T} / \mathrm{s}$$

From the "kink" in the graph of Figure , we conclude that the radius of the circular region is $$2.0 \mathrm{cm} .$$ For values of $$r$$ less than that, we have (from the absolute value of Eq. $$\oint \vec{E} \cdot d\vec{s}=-\dfrac{d\Phi _B}{dt} \text{ (Faraday’s law) }$$ )

$$E(2 \pi r)=\dfrac{d \Phi_{B}}{d t}=\dfrac{d(B A)}{d t}=A \dfrac{d B}{d t}=\pi r^{2} a$$

which
means that $$E / r=a / 2 .$$ This corresponds to the slope of that graph (the
linear portion for small values of $$r$$ ) which we estimate to be 0.015 (in SI
units). Thus, $$a=0.030 \mathrm{T} / \mathrm{s}$$

(a) The point at which we are evaluating the field is inside the solenoid, so Eq. $$E=\dfrac{r}{2}\dfrac{dB}{dt}$$
applies. The magnitude of the induced electric field is

$$E=\dfrac{1}{2} \dfrac{d B}{d t} r=\dfrac{1}{2}\left(6.5 \times 10^{-3} \mathrm{T} / \mathrm{s}\right)(0.0220 \mathrm{m})=7.15 \times 10^{-5} \mathrm{V} / \mathrm{m}$$

(b) Now
the point at which we are evaluating the field is outside the solenoid and Eq.
$$E=\dfrac{R^2}{2r} \dfrac{dB}{dt}$$ applies. The magnitude of the induced
field is

$$E=\dfrac{1}{2} \dfrac{d B}{d t} \dfrac{R^{2}}{r}=\dfrac{1}{2}\left(6.5 \times 10^{-3} \mathrm{T} / \mathrm{s}\right) \dfrac{(0.0600 \mathrm{m})^{2}}{(0.0820 \mathrm{m})}=1.43 \times 10^{-4} \mathrm{V} / \mathrm{m}$$

(a) It should be emphasized that the result, given in terms of $$\sin (2 \pi f t)$$ , could as easily be given in terms of $$\cos (2 \pi f t)$$ or even $$\cos (2 \pi f t+\phi)$$ where $$\phi$$ is a phase constant . The angular position $$\theta$$ of the rotating coil is measured from some reference line (or plane), and which line one chooses will affect whether the magnetic flux should be written as $$B A \cos \theta, B A \sin \theta$$ or $$B A \cos (\theta+\phi) .$$ Here our choice is such that $$\Phi_{B}=B A \cos \theta .$$ since the coil is rotating steadily, $$\theta$$ increases linearly with time. Thus, $$\theta=\omega t$$ (equivalent to $$\theta=2 \pi f t$$ ) if $$\theta$$ is understood to be in radians (and $$\omega$$ would be the angular velocity). since the area of the rectangular coil is $$A=a b$$ ,
Faraday's law leads to

$$\varepsilon=-N \dfrac{d(B A \cos \theta)}{d t}=-N B A \dfrac{d \cos (2 \pi f t)}{d t}=N B a b 2 \pi f \sin (2 \pi f t)$$

which is the desired result, shown in the problem statement. The second way this is written $$\left(a_{0} \sin (2 \pi f)\right)$$ is meant to emphasize that the voltage output is sinusoidal (in its time dependence) and has an amplitude of $$\varepsilon_{0}=2 \pi f N a b B$$

(b) We solve

$$\varepsilon_{0}=150 \mathrm{V}=2 \pi f \mathrm{NabB}$$

when $$f=60.0$$ rev/s and $$B=0.500$$
T. The three unknowns are $$N,$$ a, and $$b$$ which occur in a product; thus,
we obtain $$N a b=0.796 \mathrm{m}^{2}$$

Conductor $$AB$$ moves towards right with speed 

In absolute value, Faraday's law (for a single turn, with $$B$$ changing in time) gives

$$\dfrac{d \Phi_{B}}{d t}=\dfrac{d(B A)}{d t}=A \dfrac{d B}{d t}=\pi R^{2} \dfrac{d B}{d t}$$

for the magnitude of the induced emf. Dividing it by $$R^{2}$$ then allows us to relate this to the slope of the graph in Figure b [particularly the first part of the graph], which we estimate to be $$80 \mu \mathrm{V} / \mathrm{m}^{2}$$

(a) Thus, $$\dfrac{d B_{1}}{d t}=\left(80 \mu \mathrm{V} / \mathrm{m}^{2}\right) / \pi \approx 25 \mu \mathrm{T} / \mathrm{s}$$

(b) Similar reasoning for region 2 (corresponding to the slope of the second part of the graph in Figure b) leads to an emf equal to

$$\pi r_{1}^{2}\left(\dfrac{d B_{1}}{d t}-\dfrac{d B_{2}}{d t}\right)+\pi R^{2} \dfrac{d B_{2}}{d t}$$

which means the second slope (which we estimate to be $$40 \mu \mathrm{V} / \mathrm{m}^{2}$$ ) is equal to $$\pi \dfrac{d B_{2}}{d t}$$

$$\text { Therefore, } \dfrac{d B_{2}}{d t}=\left(40 \mu \mathrm{V} / \mathrm{m}^{2}\right) / \pi \approx 13 \mu \mathrm{T} / \mathrm{s}$$

(c) Considerations of Lenz's law leads to the conclusion that $$B_{2}$$ is
increasing.

Induced current $$I$$ in loop $$ABYX = I_{i} = \dfrac {\epsilon}{R} = \dfrac {-1}{R} \dfrac {dQ}{dt}$$
$$I_{i} = \dfrac {1}{R}\cdot \dfrac {d}{dt} \vec {B} . \vec {A}$$
$$I_{t} = \dfrac {vBd}{R}$$
Angle between $$\vec {B}$$ and $$\vec {A}$$ is zero. Directin of $$I$$ from $$A$$ to $$B$$ is given in Fleming's right hand rule.
As switch $$S$$ is closed at $$t = 0$$ .
Current of first question will charge the capacitor. Let $$Q(t)$$ be the charge on capacitor.
$$\therefore Q (t) = C.V$$
$$Q(t) = C.I_{C}.R$$
$$\therefore I_{c} = \dfrac {Q(t)}{RC}$$
The capacitor opposes the flow of charge, so net current in circuit
$$I = I_{i} - I_{c}$$
$$I = \dfrac {Bvd}{R} - \dfrac {Q(t)}{RC}$$
$$\dfrac {dQ(t)}{dt} = \dfrac {Bvd}{R} - \dfrac {Q(t)}{RC}$$
$$\dfrac {Bvd}{R} = \dfrac {dQ(t)}{dt} + \dfrac {Q(t)}{RC}$$
Or $$\dfrac {dQ(t)}{dt} + \dfrac {Q(t)}{RC} =\dfrac {Bvd}{R}$$
So the solution of linear differential equation is
$$Q(t) = \dfrac {\dfrac {Bvd}{R}}{\dfrac {1}{RC}} + Ae^{\dfrac {-t}{RC}}$$
$$Q(t) = Bvd C + Ae^{-t/RC}$$
Initially at $$t = 0, Q = 0$$
So $$A = -Bvd\ C$$
So $$Q(t) = BvdC - Bvd\ C e^{\dfrac {t}{RC}} [\because e^{0} = 1]$$
$$Q(t) = BvdC \left [1 - e^{-\dfrac {t}{RC}}\right ]$$
Current in circuit $$I = \dfrac {dQ(t)}{dt} = \dfrac {Bvd}{R} e^{-\dfrac {1}{RC}}$$  

The relation $$\overrightarrow{\mu_1}=(-e/2m)\overrightarrow{L}$$
is in accordance with the classic physics. It can be derived classical as beloe:
$$\mu_1=IA\quad \quad (1)$$
If $$-e$$ is charge on electron and $$T$$ , its period of revolution, then
$$I=\frac{-e}{T}$$
Also, $$A=\pi r^2$$
Therefore, the equation $$(1)$$ becomes 
$$\mu_1= \left( \frac { -e }{ T }  \right) \times \pi r^2 \quad .....(2)$$
If $$V$$ is orbital velocity of the electron, then its orbital angular momentum is given by
$$L=m \quad v \quad r$$
= $$m\times \frac{2\pi r}{T}\times r$$
$$T=\frac{2 \pi mr^2}{L}$$
In the equation $$(2)$$ substituting for $$T$$ , we have,
$$ \
mu_1=-(e/2m)L $$
In vector notation,
$$\overrightarrow{\mu}_1=-(e/2m)\overrightarrow{L}$$
The negative sign accounts for the fact that vectors. $$\overrightarrow{\mu}$$ . and $$\overrightarrow{r}$$ are directed opposite to each other.

We know that the torque acting on a magnetic dipole.
$$\vec {N} = \vec {p_{m}}\times \vec {B}$$
But, $$\vec {p_{m}} = iS\hat {n}$$ , where $$\hat {n}$$ is the normal on the plane of the loop and is direction in the direction of advancement of a right handed screw, if we rotate the screw in the sense of current in the loop.
On passing a current through the coil, this torque acting on the magnetic dipol, is counterbalanced by the moment of additional weight, about $$O$$ . Hence, the direction of current in the loop must be in the direction, shown in figure.
$$\vec {p_{m}} \times \vec {B} = -\vec {l}\times \triangle m\vec {g}$$
or, $$N\ i\ S\ B = \triangle mgl$$
So, $$B = \dfrac {\triangle mgl}{N\ i\ S} = 0.4\ T$$ on putting the values.

In $$K, \vec {E} = a\dfrac {vec {r}}{r^{2}}, \vec {r} = (x\hat {i} + y\hat {j})$$
In $$K', \vec {B'} = -\dfrac {\vec {v}\times \vec {E}}{c^{2}} = \dfrac {a\vec {r} \times \vec {v}}{c^{2}r^{2}}$$
The magnetic lines are circular.