Electromagnetic Induction - Class 12 Medical Physics - Extra Questions

We know that,

Self-inductance of a solenoid $$=\dfrac{\mu A{{n}^{2}}}{l}$$

So, self-induction proportional to $${{n}^{2}}$$

So, induction becomes $$4$$ times when $$n$$ is doubled.

Hence, this is the required solution

(a) From Eq. $$\mathscr{E}
_L=-L\dfrac{di}{dt}\text{ (self-induced current) }$$, we find $$L=(3.00
\mathrm{mV}) /(5.00 \mathrm{A} / \mathrm{s})=0.600 \mathrm{mH}$$

(b) since $$N
\Phi=i L \text { (where } \Phi=40.0 \mu \mathrm{Wb} \text { and } i=8.00
\mathrm{A}),$$ we obtain $$N=120$$

The Faraday's laws of electromagnetic induction says that the E.M.F. induced in a coil 'e' = -(rate of change of magnetic flux linkage) 

where,
the Flux linkage =number  of  turns'N' x magnetic  field'B' x area'A' x cos$$\theta$$
where,
theta is angle between magnetic field B and area A.
Theta at any instant 't'=$$(angular  \ velocity'w')\times (time \  instant't')$$. 

That is, Theta = $$w\times t$$.

E.M.F. induced in a coil 'e'=$$N\times B\times A\times w\times sinw\times t$$. 
The factors involved in the induced emf of a coil are:

• The induced e.m.f. is directly proportional to N, the total number of turns in the coil.
• The induced e.m.f. is directly proportional to A, the area of cross-section of the coil.
• The induced e.m.f. is directly proportional to B, the strength of the magnetic field in which the coil is rotating.
• The induced e.m.f. is directly proportional to 'w', the angular velocity of coil.
• The induced e.m.f. also varies with time and depends on instant 't'.
• The induced e.m.f. is maximum when plane of coil is parallel to magnetic field B and e.m.f. is zero when plane of coil is perpendicular to magnetic field B. 

Generators are useful appliances that supply electrical power during a power outage and prevent discontinuity of daily activities or disruption of business operations. Generators are available in different electrical and physical configurations for use in different applications.

An electric generator is a device that converts mechanical energy obtained from an external source into electrical energy as the output.

The generator works on the principle of electromagnetic induction discovered by Michael Faraday in 1831-32. Faraday discovered that the above flow of electric charges could be induced by moving an electrical conductor, such as a wire that contains electric charges, in a magnetic field. This movement creates a voltage difference between the two ends of the wire or electrical conductor, which in turn causes the electric charges to flow, thus generating electric current.

The emf is related to inductance by the equation, 

$$e=L\dfrac{di}{dt}$$.

Now, $$L=\dfrac{e}{\dfrac{di}{dt}}=\dfrac{200}{\dfrac{5}{0.1}}=4H$$.

Length of the rectangular wire, $$l=8cm=0.08m$$

Width of the rectangular wire, $$b=2cm=0.02m$$

Hence, area of the rectangular loop,

$$A=lb$$

$$=0.08\times 0.02$$

$$=16\times 10^{-4}m^2$$

Magnetic field strength, $$B=0.3T$$

Velocity of the loop, $$v=1cm/s=0.01m/s$$

(a)

Emf developed in the loop is given as:

$$e=Blv$$

   $$=0.3\times 0.08\times 0.01=2.4\times 10^{-4}V$$

Time taken to travel along the width, $$t\displaystyle =\frac{Distance\, travelled}{Velocity}=\frac{b}{v}$$

                                                                 $$\displaystyle =\frac{0.02}{0.01}=2s$$

Hence, the induced voltage is $$2.4\times 10^{-4}V$$ which lasts for 2 s.

(b)

Emf developed, $$e=Bbv$$

$$=0.3\times 0.02\times 0.01=0.6\times 10^{-4}V$$

Time taken to travel along the length, $$\displaystyle t=\frac{Distance\, traveled}{Velocity}=\frac{l}{v}$$

                                                                  $$=\displaystyle \frac{0.08}{0.01}=8s$$

Hence, the induced voltage is $$0.6\times 10^{-4}V$$ which lasts for 8 s.

Given that,

Number of turns $$N=60$$

Length $$l=20\,cm$$

Width $$b=10\,cm$$

Magnetic field $$B=0.5\,T$$

Rotation per min $$f=1800$$

We know that,

$$ \phi =\left( AB\cos \theta  \right)N $$

 $$ \phi =NAB\cos \omega t $$

 $$ \dfrac{-d\phi }{dt}=\varepsilon  $$

 $$ \varepsilon =-\dfrac{d\left( NAB\cos \omega t \right)}{dt} $$

 $$ \varepsilon =-AB\omega \left( -\sin \omega t \right)N $$

 $$ \varepsilon =NAB\omega \sin \omega t $$

 $$ \omega =2\pi f $$

 $$ \varepsilon =2\pi fNAB\sin \omega t $$

 $$ {{\varepsilon }_{\max }}=2\pi fNAB $$

 $$ {{\varepsilon }_{\max }}=2\times 3.14\times 60\times \left( 20\times 10\times {{10}^{-4}} \right)\times 0.5\times 30 $$

 $$ {{\varepsilon }_{\max }}=113.04\,volt $$

Hence, the maximum induced emf is $$113.04\,volt$$ 

Rotating dics can be consider as a rotating rod of same radius.

v at a distance $$r/2$$ From the centre $$= r \dfrac{w}{2}$$

$$E=Blv$$

$$E= B \times r \times \dfrac{rw}{2}$$

$$ E= \dfrac{1}{2} B r^2 w$$ 

• Rectangular coil: Rectangular Coil is the conductor coil in which current is produced. When the coil rotates in an applied magnetic field due to electromagnetic induction a current is produced.
• Permanent magnet: permanent magnets are another most important part. They provide a steady magnetic field in which the coil rotates. Without them there is no magnetic field and there will be no electromagnetic induction.
• Slip rings: When the coil rotated, the wire attached to them through which we will get electricity will also rotate and will create a problem by as the wire will get a curl up. So the solution to this problem is slip rings. The current which is produced through coil is transported to the rings. Inner part of the ring is allowed to have rotation but the exterior part is not allowed and inner and exterior part are both connected. So the current passes to exterior part from inner part and then to the wire.
• Carbon brushes: Carbon brushes are the fixed part which is connected to slip rings. They are fixed and current from slip rings passes to carbon brush and then to the wire.
• Shaft: Shaft is like a handle attached to the coil, which is used for moving the coil. It basically makes rotation easier and faster.

Since
$$\varepsilon =-L(di/dt)$$, we may obtain the desired induced emf by setting

$$\dfrac{d
i}{d t}=-\dfrac{\varepsilon}{L}=-\dfrac{60 \mathrm{V}}{12 \mathrm{H}}=-5.0
\mathrm{A} / \mathrm{s}$$

or $$|d i
/ d t|=5.0 \mathrm{A} / \mathrm{s} .$$ We might, for example, uniformly reduce
the current from $$2.0 \mathrm{A}$$ to zero in $$40 \mathrm{ms}$$

The magnetic field $$B$$ can be expressed as

$$B(t)=B_{0}+B_{1}
\sin \left(\omega t+\phi_{0}\right)$$

where $$B_{0}=(30.0
\mathrm{T}+29.6 \mathrm{T}) / 2=29.8 \mathrm{T}$$ and $$B_{1}=(30.0
\mathrm{T}-29.6 \mathrm{T}) / 2=0.200 \mathrm{T} .$$ Then

from Eq. $$E=\dfrac{r}{2}\dfrac{dB}{dt}$$

$$E=\dfrac{1}{2}\left(\dfrac{d
B}{d t}\right) r=\dfrac{r}{2} \dfrac{d}{d t}\left[B_{0}+B_{1} \sin \left(\omega
t+\phi_{0}\right)\right]=\dfrac{1}{2} B_{1} \omega r \cos \left(\omega
t+\phi_{0}\right)$$

We note
that $$\omega=2 \pi f$$ and that the factor in front of the cosine is the
maximum value of the field. Consequently,

$$E_{\max
}=\dfrac{1}{2} B_{1}(2 \pi f) r=\dfrac{1}{2}(0.200 \mathrm{T})(2 \pi)(15
\mathrm{Hz})\left(1.6 \times 10^{-2} \mathrm{m}\right)=0.15 \mathrm{V} /
\mathrm{m}$$

(a) The self-inductance per meter is

$$\dfrac{L}{\ell}=\mu_{0} n^{2}
A=\left(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)(100 \mathrm{turns}
/ \mathrm{cm})^{2}(\pi)(1.6 \mathrm{cm})^{2}=0.10 \mathrm{H} / \mathrm{m}$$

(b) The induced emf per meter is

$$\dfrac{\varepsilon}{\ell}=\dfrac{L}{\ell}
\dfrac{d i}{d t}=(0.10 \mathrm{H} / \mathrm{m})(13 \mathrm{A} / \mathrm{s})=1.3
\mathrm{V} / \mathrm{m}$$

(a) Equation $$\mathscr{E}=\dfrac{d\Phi
_B}{dt}=\dfrac{d}{dt}BLx=BL\dfrac{dx}{dt}=BLv$$ leads to

$$\varepsilon=B L v=(0.350
\mathrm{T})(0.250 \mathrm{m})(0.55 \mathrm{m} / \mathrm{s})=0.0481 \mathrm{V}$$

(b) By Ohm's law, the induced current is

$$i=0.0481 \mathrm{V} / 18.0
\Omega=0.00267 \mathrm{A}$$

By Lenz's law, the current is clockwise in Figure.

(c) Equation $$P=i^2R\text{ (resistive dissipation) }$$ leads to $$P=i^{2}
R=0.000129 \mathrm{W}$$

The total induced emf is given by

$$\varepsilon =-N \dfrac{d
\Phi_{B}}{d t}=-N A\left(\dfrac{d B}{d t}\right)=-N A \dfrac{d}{d
t}\left(\mu_{0} n i\right)=-N \mu_{0} n A \dfrac{d i}{d t}=-N \mu_{0}
n\left(\pi r^{2}\right) \dfrac{d i}{d t} $$

$$=-(120)\left(4 \pi \times 10^{-7}
\mathrm{T} \cdot \mathrm{m} / \mathrm{A}\right)(22000 / \mathrm{m}) \pi(0.016
\mathrm{m})^{2}\left(\dfrac{1.5 \mathrm{A}}{0.025 \mathrm{s}}\right) $$

$$=0.16 \mathrm{V}$$

Ohm's law then yields $$i=|\varepsilon|
/ R=0.016 \mathrm{V} / 5.3 \Omega=0.030 \mathrm{A}$$

(a) Immediately after the switch is closed, $$\varepsilon-\varepsilon_{L}=i R .$$
But $$i=0$$ at this instant, so $$\varepsilon_{L}=\varepsilon,$$ or $$\varepsilon_{L}
/ \varepsilon=1.00$$

(b) $$\varepsilon_{L}(t)=\varepsilon
e^{-t / \tau_{L}}=\varepsilon e^{-2.0 \tau_{L} / \tau_{L}}=\varepsilon
e^{-2.0}=0.135 \varepsilon,$$ or $$\varepsilon_{L} / \varepsilon=0.135$$

(c) From $$\varepsilon_{L}(t)=\varepsilon
e^{-t / \tau_{L}}$$ we obtain

$$\dfrac{t}{\tau_{L}}=\ln
\left(\dfrac{\varepsilon}{\varepsilon_{L}}\right)=\ln 2 \Rightarrow t=\tau_{L}
\ln 2=0.693 \tau_{L} \Rightarrow t / \tau_{L}=0.693$$

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

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

In this problem, we find $$\dfrac{d B}{d t}=-\dfrac{B_{0}}{\tau} e^{-t l \tau} .$$
Thus, $$\varepsilon=\pi r^{2} \dfrac{B_{0}}{\tau} e^{-t / \tau}$$

Given: current in solenoids falls from 6 A to 0 in just 2.5 ms, emf=30 kV

We use $$\varepsilon_{2}=-M
d i_{1} / d t \approx M \Delta i / \Delta t |$$ to find $$M$$

$$M=\left|\dfrac{\varepsilon}{\Delta
i_{1} / \Delta t}\right|=\dfrac{30 \times 10^{3} \mathrm{V}}{6.0 \mathrm{A}
/\left(2.5 \times 10^{-3} \mathrm{s}\right)}=13 \mathrm{H}$$

The induced electric field is given by Eq. $$\oint \vec{E} \cdot d\vec{s}=-\dfrac{d\Phi
_B}{dt}\text{ (Faraday’s Law) }$$:

$$\oint
\vec{E} \cdot d \vec{s}=-\dfrac{d \Phi_{B}}{d t}$$

The electric field lines are circles that are concentric with the cylindrical
region. Thus,

$$E(2 \pi
r)=-\left(\pi r^{2}\right) \dfrac{d B}{d t} \Rightarrow E=-\dfrac{1}{2} \dfrac{d
B}{d t} r$$

The force
on the electron is $$\vec{F}=-e \vec{E}$$, so by Newton's second law, the acceleration is

$$\vec{a}=-e
\vec{E} / m$$

(a) At point $$a$$

$$E=-\dfrac{r}{2}\left(\dfrac{d
B}{d t}\right)=-\dfrac{1}{2}\left(5.0 \times 10^{-2} \mathrm{m}\right)\left(-10
\times 10^{-3} \mathrm{T} / \mathrm{s}\right)=2.5 \times 10^{-4} \mathrm{V} /
\mathrm{m}$$

With the normal taken to be into the page, in the direction of the magnetic field, the positive direction for $$\vec{E}$$ is clockwise. Thus, the direction of the electric field at point a is to the left, that is $$\vec{E}=-\left(2.5 \times
10^{-4} \mathrm{V} / \mathrm{m}\right)$$ i. The resulting acceleration is

$$\vec{a}_{a}=\dfrac{-e
\vec{E}}{m}=\dfrac{\left(-1.60 \times 10^{-19} \mathrm{C}\right)\left(-2.5
\times 10^{-4} \mathrm{V} / \mathrm{m}\right)}{\left(9.11 \times 10^{-31}
\mathrm{kg}\right)} \hat{\mathrm{i}}=\left(4.4 \times 10^{7} \mathrm{m} /
\mathrm{s}^{2}\right) \hat{\mathrm{i}}$$

The acceleration is to the right.

(b) At point $$b$$ we have $$r_{b}=0,$$ so the acceleration is zero.

(c) The electric field at point $$c$$ has the same magnitude as the field in $$a$$, but
with its direction reversed. Thus, the acceleration of the electron released at
point $$c$$ is

$$\vec{a}_{c}=-\vec{a}_{a}=-\left(4.4
\times 10^{7} \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}$$

Equation $$P=\dfrac{V^2}{R}$$
gives $$\varepsilon^{2} / R$$ as the rate of energy transfer into thermal forms $$\left(d E_{\mathrm{th}} / d t\right.$$ which, from Figure c , is roughly $$40
\mathrm{nJ} / \mathrm{s}$$ ). Interpreting $$\varepsilon$$ as the induced emf (in absolute value) in the single-turn loop $$(N=1)$$ from Faraday's law, we have

$$\varepsilon=\dfrac{d
\Phi_{B}}{d t}=\dfrac{d(B A)}{d t}=A \dfrac{d B}{d t}$$

Equation $$B=\mu _0in$$ gives $$B=\mu_{0} n i$$ for the solenoid (and note that the field is zero outside of the solenoid, which implies that $$A=A_{\text {coil }}$$, so our expression for the magnitude of the induced emf becomes

$$\varepsilon=A
\dfrac{d B}{d t}=A_{\text {coil }} \dfrac{d}{d t}\left(\mu_{0} n_{\text { coil }}\right)=\mu_{0}
n A_{\text { coil }} \dfrac{d i_{\text {coil }}}{d t}$$

where Figure b, suggests that $$d_{\mathrm{ coil }} / d t=0.5 \mathrm{A} / \mathrm{s} .$$ With $$n=8000$$ (in SI units) and $$A_{\mathrm{coil}}$$ $$=\pi(0.02)^{2}$$ (note that the loop radius does not come into the computations of this problem, just the coil's), we find $$\mathrm{V}=6.3 \mu \mathrm{V}$$. Returning to our earlier observations, we can now solve for the resistance:

$$R=\varepsilon^{2}
/\left(d E_{\mathrm{th}} / d t\right)=1.0 \mathrm{m} \Omega$$