State the principle of an ac generators and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil havin N turns each of cross-sectional area, rotating with a constant angular speed 'ω' in a magnetic field $\overrightarrow{\mathrm{B}}$, directed perpendicular to the axis of rotation.
Principle − AC generator based on the phenomenon of electromagnetic induction.
Construction:
Main parts of an ac generator:
Armature − Rectangular coil ABCD
Filed Magnets − Two pole pieces of a strong electromagnet
Slip Rings − The ends of coil ABCD are connected to two hollow metallic rings R_{1} and R_{2}.
Brushes − B_{1} and B_{2} are two flexible metal plates or carbon rods. They are fixed and are kept in tight contact with R_{1} and R_{2} respectively.
Theory and Working − As the armature coil is rotated in the magnetic field, angle θ between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An emf is induced in the coil. According to Fleming’s right-hand rule, current induced in AB is from A to B and it is from C to D in CD. In the external circuit, current flows from B_{2} to B_{1}.
To calculate the magnitude of emf induced:
Suppose
A → Area of each turn of the coil
N → Number of turns in the coil
$\overrightarrow{\mathrm{B}}$→ Strength of magnetic field
θ → Angle which normal to the coil makes with at any instant t
∴ Magnetic flux linked with the coil in this position:
$\mathrm{\Phi}=\mathrm{N}(\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{A}})$= NBA cosθ= NBA cosωt …(i)
Where, ‘ω’ is angular velocity of the coil
As the coil rotates, angle θchanges. Therefore, magnetic flux Φ linked with the coil changes and hence, an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then
$\mathrm{e}=-\frac{\mathrm{d\theta}}{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{NAB}\mathrm{cos}\mathrm{\omega t})\phantom{\rule{0ex}{0ex}}=-\mathrm{NAB}\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{cos}\mathrm{\omega t})\phantom{\rule{0ex}{0ex}}=-\mathrm{NAB}(-\mathrm{sin}\mathrm{\omega t})\mathrm{\omega}$
A device X is connected across an ac source of voltage V=V_{0}sin ωt. The current through X is given as $\mathrm{I}-{\mathrm{I}}_{0}\mathrm{sin}\left(\mathrm{\omega t}+\frac{\mathrm{\pi}}{2}\right)$
Define electric flux. Is it a scalar or a vector quantity?
A point charge q is at a distance of d/2 directly above the centre of a square of side d, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.
b) If the point charge is now moved to a distance 'd' from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.
The total number of electric field lines crossing an area placed normal to the electric field is termed as electric flux.
It is denoted by Φ.
${\mathrm{\varphi}}_{\mathrm{E}}=\overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{ds}}$
Electric flux is a scalar quantity, its SI unit is Nm^{2}C^{-1}
Electric flux through square is
${\mathrm{\varphi}}_{\mathrm{E}}=\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{0}6}$
b) Flux will not be changed,
i.e., ${\mathrm{\varphi}}_{\mathrm{E}}=\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{0}6}$
An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10^{–4} T and the angle of dip is 30°.
e = B_{v}.V.l
=$5x{10}^{-4}\mathrm{sin}30x900x\frac{5}{18}x20\phantom{\rule{0ex}{0ex}}=1.25V$
a) Use Gauss’ law to derive the expression for the electric field $\overrightarrow{E}$ due to a straight uniformly charged infinite line of charge density λ C/m.
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r_{1} to r_{2} (r_{2}>r_{1}).
Gauss’s law in electrostatics: It states that total electric flux over the closed surface S is $\frac{1}{{\epsilon}_{0}}$times the total charge (q) contained inside S.
$\therefore \overline{)\underset{\mathrm{S}}{\oint}\overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{ds}}=\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{0}}}$
Electric field due to an infinitely long straight wire.
Electric field due to an infinitely long straight wire.
Let us consider an infinitely long line charge having linear charge density λ. Assume a cylindrical Gaussian surface of radius r and length 1 coaxial with the line charge.
By symmetry, the electric field E has the same magnitude at each point of the curved surface S_{1} and is directed radially outward. So angle at surfaces between $\overrightarrow{{\mathrm{dS}}_{1}}\mathrm{and}{\overrightarrow{\mathrm{E}}}_{1}$ is zero, and angle of $\overrightarrow{{\mathrm{dS}}_{2}},\overrightarrow{{\mathrm{dS}}_{3}}\mathrm{with}\overrightarrow{\mathrm{E}}\mathrm{at}{\mathrm{S}}_{2}\mathrm{and}{\mathrm{S}}_{3}\mathrm{are}90\xb0$
Total flux through the cylindrical surface,
$\oint \overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{ds}}=\underset{{\mathrm{S}}_{1}}{\oint}\overrightarrow{\mathrm{E}}.\overrightarrow{{\mathrm{dS}}_{1}}+\underset{{\mathrm{S}}_{2}}{\oint}\overrightarrow{\mathrm{E}}.\overrightarrow{{\mathrm{dS}}_{2}}+\underset{{\mathrm{S}}_{3}}{\oint}\overrightarrow{\mathrm{E}}.\overrightarrow{{\mathrm{dS}}_{3}}\phantom{\rule{0ex}{0ex}}\underset{{\mathrm{S}}_{1}}{\oint}{\mathrm{EdS}}_{1}.\mathrm{cos}0\xb0+\underset{{\mathrm{S}}_{2}}{\oint}{\mathrm{EdS}}_{2}.\mathrm{cos}90\xb0+\underset{{\mathrm{S}}_{3}}{\oint}{\mathrm{EdS}}_{3}.\mathrm{cos}0\xb0\phantom{\rule{0ex}{0ex}}=\mathrm{E}\oint {\mathrm{dS}}_{1}=\mathrm{E}\mathrm{x}2\mathrm{\pi rl}$
Since λ is the charge per unit length and l is the length of the wire.
Thus, the charge enclosed
q = λl
According to Gaussian law,
$\oint \overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{dS}}=\frac{\mathrm{q}}{{\mathrm{\epsilon}}_{0}}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{E}\mathrm{x}2\mathrm{\pi rl}=\frac{\mathrm{\lambda l}}{{\mathrm{\epsilon}}_{0}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{E}=\frac{\mathrm{\lambda}}{2\mathrm{\pi}{\mathrm{\epsilon}}_{0}\mathrm{r}}$
b)
c)
$\mathrm{Work}\mathrm{done}=-{\int}_{{\mathrm{r}}_{1}}^{{\mathrm{r}}_{2}}\mathrm{q}\mathrm{E}.\mathrm{dr}\phantom{\rule{0ex}{0ex}}=\frac{-\mathrm{q\lambda}}{2{\mathrm{\pi \epsilon}}_{0}}\mathrm{log}\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}$
Four nuclei of an element undergo fusion to form a heavier nucleus, with the release of energy. Which of the two — the parent or the daughter nucleus — would have higher binding energy per nucleon?
Daughter nuclei are more stable than parent nuclei.
Draw graphs showing a variation of the photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.
Name the electromagnetic radiations used for (a) water purification, and (b) eye surgery
(a) Electromagnetic radiation used for water purification: U.V (Ultra Violet) rays.
(b) Electromagnetic radiation used for eye surgery: Infrared Radiation (Laser light).
The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lower voltage.
A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency ?
Mass of electron is low as compared to proton. Hence when both enter into the uniform magnetic region, the electron will move in a circular path with higher frequency on the opposite direction to the current.