Advertisements
Advertisements
प्रश्न
Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of
(i) Magnetic flux and
(ii) Alternating emf versus time generated by a loop of wire rotating in a magnetic field.
Advertisements
उत्तर
Principle − Based on the phenomenon of electromagnetic induction.
Construction:
Main parts of an ac generator:
Armature − The rectangular coil ABCD
Filed Magnets − Two pole pieces of a strong electromagnet
Slip Rings − The ends of the coil ABCD are connected to two hollow metallic rings R1 and R2.
Brushes − B1 and B2 are two flexible metal plates or carbon rods. They are fixed and are kept in tight contact with R1 and R2, respectively.
Working − As the armature coil is rotated in the magnetic field, angle θ between the field and the normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes and an emf is induced in the coil. According to Fleming’s right hand rule, current is induced from A to B in AB and from C to D in CD. In the external circuit, current flows from B2 to B1.
To calculate the magnitude of emf induced:
Suppose A → Area of each turn of the coil
N → Number of turns in the coil
`vecB`→ Strength of the magnetic field
θ → Angle which normal to the coil makes with `vecB` at any instant t
∴ Magnetic flux linked with the coil in this position is given by,
`phi=N(vecB.vecA)=NBAcostheta=NBAcos `
Where, ‘ω’ is angular velocity of the coil
Graph between magnetic flux and time, according to equation (i), is shown below:
As the coil rotates, angle θ changes. Therefore, magnetic flux Φ linked with the coil changes and an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then
`e=(dtheta)/dt=-d/dt(NABcos`
`=-NABd/dt(cos `
=-NAB(-sin ωt)ω
∴ e = NAB ω sin ωt
The graph between alternating emf versus time is shown below:
APPEARS IN
संबंधित प्रश्न
Ram is a student of class X in a village school. His uncle gifted him a bicycle with a dynamo fitted in it. He was very excited to get it. While cycling during night, he could light the bulb and see the objects on the road. He, however, did not know how this device works. he asked this question to his teacher. The teacher considered it an opportunity to explain the working to the whole class.
Answer the following questions:
(a) State the principle and working of a dynamo.
(b) Write two values each displayed by Ram and his school teacher.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
A rectangular coil having 60 turns and area of 0.4m2 is held at right angles to a uniform magnetic field of flux density 5 × 10-5T. Calculate the magnetic flux passing through it.
Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a velocity v towards the arm Rs. Assuming that the arms QR, RS and SP have negligible resistances and the moving arm PQ has the resistance r, obtain the expression for (i) the current in the loop (ii) the force and (iii) the power required to move the arm PQ.
Figure shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will __________ .
The magnetic flux linked with a coil in Wb is given by the equation Φ = 3t2 + 4t + 9. Then the magnitude of induced emf at t = 2 sec will be ______.
A loop, made of straight edges has six corners at A(0, 0, 0), B(L, O, 0) C(L, L, 0), D(0, L, 0) E(0, L, L) and F(0, 0, L). A magnetic field `B = B_o(hati + hatk)`T is present in the region. The flux passing through the loop ABCDEFA (in that order) is ______.
Consider a closed loop C in a magnetic field (Figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula φ = B1.dA1 + B2.dA2 +... Now if we chose two different surfaces S1 and S2 having C as their edge, would we get the same answer for flux. Jusity your answer.
A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be: (Assume the coil to be short circuited.)
In a coil of resistance 100 Ω a current is induced by changing the magnetic flux through it. The variation of current with time is shown in the figure. The magnitude of change in flux through the coil is ______.
