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What is Motional Electromotive Force?
Let us assume a uniform magnetic field \(\vec{B}\) directed along y – axis which is vertically downwards into the page [Fig.],
A rod PQ parallel to z -axis, is moving at a constant velocity \(\vec{v}\) along x -axis through this field.
In this condition, a free charge q in the rod experiences a magnetic force, \(\vec{F}_m\) = \(\overrightarrow{q v} \times \vec{B}\). If q is a positive charge then this force is directed from Q to P (i.e., along +ve z -axis). On the other hand, if q is negative then this force is directed from P to Q (i.e., along -ve z-axis) [Fig.]. Hence the free charges continue to accumulate at the ends of the rod creating a gradually increasing potential difference between P and Q.
This in turn creates an increasing electric field \(\vec{E}\) within the rod, in the direction from P to Q (opposite to the magnetic force). Due to this growing up electric field, the electric force exerted on the charge q at any moment, Fe = qE.
As the magnitude of electric force comes to be equal with that of magnetic force, no net force anymore does act on the charge q, i.e., at equilibrium condition,
Let the potential difference between the ends P and Q be V. If the length of the rod is l then,
V = El = vBl [putting the value of E from (1)] ……. (2)
Now, if the ends of the rod are connected with a conductive wire, a current is set up in the circuit [Fig.], Hence the moving rod can be considered as a source of emf (e), i.e., as a cell. Just like in a cell, this emf directs from the negative end Q to the positive end P inside the rod.
For an open circuit, equation (2) can be written as,
V = e = vBl ……. (3)
This induced emf in the moving rod is called motional electromotive force.
In general, if θ be the angle between \(\vec{v}\) and \(\vec{B}\) then, equation (3) becomes,
e = vBlsinθ …… (4)
Discussions:
Dependence of induced motional emf on different factors: From the above equation (4), the emf e induced across the straight conductor moving in a magnetic field is—
i) directly proportional to the magnetic induction (B),
ii) directly proportional to the length (l) of the conductor,
iii) directly proportional to the velocity (v) of the conductor and
iv) dependent on the angle θ between the magnetic field and the direction of motion of the conductor. If the direction of motion of the conductor is parallel to the magnetic field then, θ = 0 and e = 0, and in that case, no electromagnetic induction takes place.
For θ = 90° , the emf induced is maximum.
Emf induced between the two extremities of the wings of an aeroplane: At different places on the earth, the geomagnetic field is not horizontal; from the values of angle of dip, the actual direction of the geomagnetic field is known. When an aeroplane flies in a horizontal plane above earth’s surface, the length of its two wings repeatedly intercepts the lines of force of the geomagnetic field. As a result, electromagnetic induction takes place, i.e., a potential difference is set up between the two extremities of its wings. The magnitude of this potential difference depends on
- the distance between the extremities of the wings,
- the velocity of the aeroplane,
- the direction of motion of the aeroplane,
- the horizontal component and
- the angle of dip of the geomagnetic field.
The value of the angle of dip at the geomagnetic equator is zero. In this position, a horizontally moving aeroplane does not cut the geomagnetic lines of force, and hence no potential difference is set up between the extremities of the wings.
Emf induced between the two ends of a conductor rotating with uniform angular velocity in a uniform magnetic field: Let a conductor of length L be rotating about the point O with uniform angular speed ω in the plane of the paper [Fig.]. The magnetic field B is normally upwards relative to the plane of the paper.
Let us consider a small element dx of the conductor at a distance x from the point O, which is moving with velocity v perpendicular to the direction of the magnetic field. The emf induced in this element of the conductor,
de = B(dx)v [using the relation, e = Blv]
= Bωxdx [∵ v = ωx]
So, the emf induced between the two ends of the entire conductor,
e = \(\int_0^L B \omega x d x\) = \(B \omega\left[\frac{x^2}{2}\right]_0^L\) = \(\frac{1}{2} B \omega L^2\)
Induced current in a moving straight conductor: In the discussion of motional electromotive force, if R be the total resistance of the circuit then, V = IR . So, from (3) we get,
IR = vBl or, I = \(\frac{1}{2} B \omega L^2\) … (5)
The direction of the induced current can be determined with the help of the following simple rule.
Fleming’s right hand rule: The thumb, the forefinger and the middle finger of the right hand are stretched perpendicular to each other. If the forefinger points in the direction of the magnetic field and the thumb in the direction of the motion of the conductor, then the middle finger will point in the direction of the induced current [Fig.].
This rule is also called the dynamo rule.
Numerical Examples
Example 1.
The distance between the two end-points of the wings of an aeroplane is 5 m and the aeroplane is flying par-allel to earth’s surface with a velocity of 360 km ᐧ h-1.
If the geomagnetic intensity is 4 × 10-4 Wb ᐧ m-2 and the angle of dip at that place is 30°, determine the emf induced between the two end-points of the wings.
Solution:
While flying horizontally, the wings of the aeroplane cuts the vertical component of earth’s magnetic field normally and hence, an emf is induced between the two ends of the wings.
This induced emf, e = Blvsinθ.
Here, B = 4 × 10-4 Wb ᐧ m-2, l = 5 m
v = 360 km ᐧ h-1 = 360 × \(\frac{5}{18}\)m ᐧ s-1 = 100 m ᐧ s-1
and θ = 30°
e = 4 × 10-4 × 5 × 100 × sin 30° = 0.1 V
Example 2.
A copper disc of diameter 20 cm is rotating uniformly about its horizontal axis passing through the centre with angular frequency 600 rpm. A uniform magnetic
field of strength 10-2 T acts perpendicular to the plane of the disc. Calculate the induced emf between its centre and a point on the rim of the disc. [WBCHSE Sample Question]
Solution:
Diameter of the disc = 20 cm
∴ Radius, r = 10 cm = 0.1 m
Angular frequency, n = 600 rpm
∴ Angular speed,
ω = \(\frac{600 \times 2 \pi \mathrm{rad}}{60 \mathrm{~s}}\) = 20π rad ᐧ s-1
Let us take a small segment dx on the disc at a distance x from its centre [Fig.]. Length of dx is so small that speed of all the points on this segment is considered to be the same, which is, v’ = ωx.
Therefore, motional emf across dx,
de = v’Bdx [where B is the magnetic field]
∴ Total induced emf between the centre and a point on the rim of the disc,
e = \(\int_0^r v^{\prime} B d x\) = \(\int_0^r \omega x B d x\) = \(\frac{1}{2} \omega B r^2\)
= \(\frac{1}{2}\) × 20π × 10-2 × (0.1)2
Example 3.
A pair of parallel horizontal conducting rails of negligi-ble resistance shorted at one end is fixed on a smooth table. The distance between the rails is L. A massless conducting rod of resistance R can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to another edge of the table. A mass m, tied to the other end of the string, hangs vertically. A constant magnetic field B exists along the perpendicular to the plane of the table in upward direction. If the system is released from rest, calculate
[i] the terminal velocity of the rod,
[ii] the acceleration of the mass at the instant, when the velocity of the rod is half the terminal velocity.
Solution:
[i] Let, rod A’B’ of lengh L and resistance R slide with a velocity v along y -axis, while a constant magnetic field B exists along z-direction.
Induced emf in the rod, e = BLv
The induced current flowing through the rod,
I = \(\frac{e}{R}\) = \(\frac{B L v}{R}\)
The direction of flow of the current through the rod A’B’ will be such that, it opposes the cause of generation of induced emf i.e., the motion of the rod along positive y -axis. Thus the rod will experience force along negative y -axis. The magnitude of the force is given by,
F = BIL = B ᐧ \(\frac{B L v}{R}\) ᐧ L = \(\frac{B^2 L^2 v}{R}\)
Let the rod move with acceleration a, along positive y -axis. Then from the equation of motion,
ma = mg – F = mg – \(\frac{B^2 L^2 V}{R}\) or, a = g – \(\frac{B^2 L^2 v}{m R}\)
When the rod gets terminal velocity v0, then a = 0
∴ 0 = g – \(\frac{B^2 L^2 v_0}{m R}\)
∴ v0 = \(\frac{m g R}{B^2 L^2}\)
[ii] When the rod moves with velocity, v = \(\frac{1}{2} v_0\), then
v = \(\frac{m g R}{2 B^2 L^2}\)
Hence the acceleration of the mass at the instant,
a = g – \(\frac{B^2 L^2}{m R} \cdot \frac{m g R}{2 B^2 L^2}\) = g – 0.5g = 0.5g