Contents
Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.
What is the Formula for Induced EMF? What do you Mean by Magnetic Flux and Magnetic Flux Density?
From Faraday’s second law we know that, if the change in magnetic flux linked with a closed conductor be dϕ in time dt,
induced emf, e ∝ \(\frac{d \phi}{d t}\)
So, according to Lenz’s law we can write,
e = -k\(\frac{d \phi}{d t}\) …. (1)
Here, k is a positive constant. Since the induced emf e opposes the change in magnetic flux dtp, a negative sign is used in the equation.
Note that, equation (1) expresses the three laws of electromagnetic induction simultaneously.
- If magnetic flux does not change i.e., if dϕ = 0 then e = 0. This supports Faraday’s first law.
- e ∝ \(\frac{d \phi}{d t}\). It is Faraday’s second law. For a coil having n turns, e ∝ \(n \frac{d \phi}{d t}\).
- The ‘negative’ sign on the right hand side of the equation (1) indicates the opposing nature of induced emf e due to the change in magnetic flux d. This is Lenz’s law.
Units of magnetic flux and magnetic induction:
The unit of magnetic flux is so defined that the magnitude of the constant k in equation (1) becomes 1.
For e = 1, if \(\frac{d \phi}{d t}\) = 1, then k = 1
In that case, e = –\(\frac{d \phi}{d t}\) …. (2)
So, the unit of magnetic liux is defined as the change in unit time of magnetic flux linked with a conducting coil, for which 1 unit emf is induced in that coil.
Again, since the magnetic flux linked with unit area is called magnetic induction or magnetic flux density, the unit of magnetic induction or magnetic flux density \(=\frac{\text { unit of magnetic flux }}{\text { unit of area }}\)
Physical Quantity | System | Unit |
Magnetic Flux | CGS SI |
Maxwell (Mx) Weber (Wb) |
Magnetic induction | CGS SI |
maxwll/centimetre2(Mx ᐧ cm-2) or Gauss G webner/metre2(Wb ᐧ m-2) or tesla(T) |
1 Wb: The change of magnetic flux linked with a coil in is, for which 1 V emf is induced in the coil, is called 1 Wb.
In other word, 1 weber is the magnetic flux which, linking a circuit of one turn, would produce in it an electromotive force of 1 volt if it were reduced to zero at a uniform rate in 1 second.
1 Mx: The change of magnetic flux linked with a coil in 1s, for which 1 abvolt (1 abvolt = 10-8 V) emf is induced in the coil, is called 1 Mx.
In other word, 1 maxwell is the magnetic flux which, linking a circuit of one turn, would produce in it an electromotive force of 1 abvolt ¡fit were reduced to zero at a uniform rate in 1 second.
Relation between different units: As 1 volt = 108 abvolt
so, 1 Wb = 108 Mx.
Again, 1 T = \(\frac{1 \mathrm{~Wb}}{1 \mathrm{~m}^2}\) = \(\frac{10^8 \mathrm{Mx}}{10^4 \mathrm{~cm}^2}\)
= 104Mx ᐧ cm-2 = 104 G
Relation between Wb and V: According to the relation,
e = \(\frac{d \phi}{d t}\),
V = \(\frac{\mathrm{Wb}}{\mathrm{s}}\) or, Wb = V ᐧ s
Amount of charge flowing through a closed circuit for induced electromotive force: Let the number of turns of any closed coil be n, the emf induced in the coil be e and the rate of change of magnetic flux linked with the coil be \(\frac{d \phi}{d t}\). Now, if the resistance of the circuit be R, the current induced in it (taking the magnitude only),
i = \(\frac{e}{R}\) = \(\frac{n}{R} \cdot \frac{d \phi}{d t}\) [∵ e = \(n \frac{d \phi}{d t}\)]
or, idt = \(\frac{n}{R} \cdot d \phi\) ……. (3)
If the initial magnetic flux linked with the coil be ϕ1 and the final magnetic flux linked be ϕ2, then integrating equation (3) we get,
\(\int_0^t i d t\) = \(\frac{n}{R} \int_{\phi_1}^{\phi_2} d \phi\) = \(\frac{n}{R}\)(ϕ2 – ϕ1)
So, the amount of charge flowing through the circuit,
q = \(\int_0^t i d t\) = \(\frac{n}{R}\)(ϕ2 – ϕ1) …. (4)
Numerical Examples
Example 1.
A coil of resistance 100 Ω having 100 turns is placed in a magnetic field. A galvanometer of resistance 400 Ω is connected in series with it. If the coil is brought from the present magnetic field to another magnetic field in \(\frac{1}{10}\)s, determine the average emf and the current. Given, the initial and final magnetic flux linked with each turn of the coil are 1 mWb and 0.2 mWb respectively.
Solution:
Change in magnetic flux for each turn = 0.2 – 1 = -0.8 mWb
So, change in magnetic flux for 100 turns
= 100 × (-0.8) mWb = -0.08 Wb
Hence, the magnitude of average emf induced
= the negative of the rate of change of magnetic flux
= –\(\left(-\frac{0.08 \mathrm{~Wb}}{\frac{1}{10} \mathrm{~s}}\right)\) = 0.8 V
The equivalent resistance of the circuit = 100 + 400 = 500 Ω.
Hence, the average induced current = \(\frac{0.8 \mathrm{~V}}{500 \Omega}[latex] = 0.0016 A = 1.6 mA.
Example 2.
A conducting wire is wound around the great circle of a spherical balloon. This circular loop can contract with the balloon. A hemispherical cross section of the balloon is shown in figure. The initial radius of the bal-loon is 0.60 m. A uniform magnetic field B = 0.25 T exists along the perpendicular to the plane of the cir-cular loop, i.e., in +y – direction.
After 5.0 × 10-2 s the balloon is deflated to a radius of 0.30 m. What will be the average emf induced in the loop during this time?
Solution:
Initially the flux linked with the conducting loop along y -axis,
ϕi = [latex]\vec{B} \cdot \overrightarrow{A_i}\) = BAi cos0° = 0.25π × 0.602
= 0.28 Wb
After deflation of the balloon, the flux linked with the conducting loop of radius 0.30m,
Example 3.
A copper wire of diameter 0.04 in. and length 50 cm is bent in the form of a circular loop. The plane of the loop is normal to a uniform magnetic field which is increasing with time at a constant rate of 100 G ᐧ s-1. What is the rate of Joule heating in the loop? [Resistivity of copper = 1.7 × 10-8Ω ᐧ m, 1 in. = 2.54 cm]
Solution:
Radius of the wire
= 0.02 in. = 0.02 × 2.54 × 10-2 m = 5.08 × 10-4 m
Area of cross section of the wire,
A1 = πr2 = π(5.08 × 10-4)2 = 81 × 10-8 m2
Resistance of the wire,