Motional Electromotive Force: Meaning, Magnetic Force, Examples

Motional Electromotive Force: Motional emf is induced in a moving electric conductor in the presence of a magnetic field. When a helicopter lands on the ground, it enters the Earth’s magnetic field. Based on faraday’s law, the blades of the helicopter acquire a potential difference. The helicopters are brought down slowly to avoid any electric shock.

When an electrical conductor is brought into a magnetic field, emf is induced in it due to its dynamic interaction with the magnetic field. This emf is known as induced emf.  Any change in magnetic flux induces an emf opposing that change. This process is known as induction. Motion is one of the major causes of induction. For example, a magnet moved toward a coil induces an emf, and a coil moved toward a magnet produces a similar emf. This section concentrates on motion in a magnetic field that is stationary relative to the Earth, producing what is loosely called motional emf.

In this article, we will learn about motional emf definition, formula and its advantages and disadvantages.

Study About EMF Of A Cell

What is Induced Current and Motional EMF?

If conducting rod moves on two parallel conducting rails, as shown in the following figure, then the phenomenon of induced emf can also be understood by the concept of generated area (The area swept of the conductor in the magnetic field, during its motion)

As shown in the figure in time \(t\) distance travelled by conductor\(=vt\)

Area generated \(A=lvt\)

Flux linked with this area \(\phi = BA = Blvt\)

Hence induced emf \(\left| e \right| = \frac{{d\phi }}{{dt}} = Bvl\) induced current \(i = \frac{e}{{R;}}i = \frac{{Bvl}}{R}\)

The direction of induced current can be found with the help of Fleming’s right-hand rule.

Fleming’s Right-Hand Rule

According to this law, if we stretch the right-hand thumb and two nearby fingers perpendicular to one another and the first finger points in the direction of the magnetic field and the thumb in the direction of motion of the conductor, then the central finger will point in the direction of the induced current. 

Note:

Here it is worthy to note that the rod \(PQ\) is acting as a source of emf, and inside a source of emf, the direction of current is from lower potential to higher potential; so the point \(P\) of the rod is at a higher potential than \(Q\) through the current in the rod \(PQ\) is from \(Q\) to \(P.\)

Magnetic Force on a Conductor

Now current is set up in circuit (conductor). As we know, when a current-carrying conductor moves in a magnetic field, it experiences a force \({F_m} = \)Bil (maximum), whose direction can be found with the help of Fleming’s left-hand rule. 

So, here conductor \(PQ\) experiences a magnetic force \({F_m} = Bil\) in the opposite direction of its motion and \({F_m} = Bil = B\left( {\frac{{Bvl}}{R}} \right)l;{F_m} = \frac{{{B^2}v{l^2}}}{R}\)

(As a result of this force \(\left( {{F_m}} \right)\) rod speed decreases as time passes.)

Note: 

1. To move the rod with uniform velocity, an external mechanical force is required, and this is \({F_{ext}} = – {F_m}\)
   \( \Rightarrow \left| {{F_{ext}}} \right| = \frac{{{B^2}vl}}{R}\)
2.  

Power Dissipated in Moving the Conductor 

For the uniform motion of the rod \(PQ,\) the rate of doing mechanical work by an external agent or mechanical power delivered by an external source is given as
\({P_{mech}} = {p_{ext}} = \frac{{dW}}{{dt}} = {F_{ext}}.v = \frac{{{B^2}v{l^2}}}{R} \times v \Rightarrow {P_{mech}} = \frac{{{B^2}{V^2}{l^2}}}{R}\)

Also, electrical power dissipated in resistance or rate of heat dissipation across the resistance is given as
\({P_{thermal}} = \frac{H}{t} = {i^2}R = {\left( {\frac{{Bvl}}{R}} \right)^2}.R;{P_{thermal}}\frac{{{B^2}{v^2}{l^2}}}{R}\)

Note: 

1. It is clear that \({P_{mech.}} = {P_{thermal.}}\) This formula is consistent with the principle of conservation of energy.

Motion of Conductor Rod in a Vertical Plane

If conducting rod is released from rest (at \(t=0\)) as shown in the figure, then with the rise in its speed \((v),\) induces emf \((e),\) induced current \((i),\) magnetic force \(\left( {{F_m}} \right)\) increases, but its weight remains constant.

Rod will achieve a constant maximum (terminal) velocity \({v_r}\) if \({F_m} = mg\)

So \(\frac{{{B^2}{V_r}^2{l^2}}}{R} = mg\)

\( \Rightarrow {v_r} = \frac{{mgR}}{{{B^2}{l^2}}}\)

Motion of Conducting Rod on an Inclined Plane

When the conductor starts sliding from the top of an inclined plane, as shown, it moves perpendicular to its length but at an angle \(\left( {90–\theta } \right)\) with the direction of a magnetic field.

Hence induced emf across the ends of a conductor

\(e = Bv\sin \left( {90 – \theta } \right)l = Bvl\cos \theta \)

So induced current \(i = \frac{{Bvl\cos \theta }}{R}\) (Directed from \(Q\) to \(P\)). 

The forces acting on the bar are shown in the following figure-

The rod will move down with constant velocity only if:

\({F_m}\cos \theta = mg\cos (90 – \theta ) = mg\sin \theta \)

\(Bil\cos \theta = mg\sin \theta \)

\(B\left( {\frac{{B{v_T}l\cos \theta }}{R}} \right)l\cos \theta = mg\sin \theta \Rightarrow {v_T} = \frac{{mgR\sin \theta }}{{{B^2}{l^2}{{\cos }^2}\theta }}\)

Note: 

The emf induced in different examples are as follows-

Movement of a train in Earth’s magnetic field: When a train moves on rails, then a potential difference between the ends of the axle of the wheels is induced because the axle of the wheels of the train cuts the vertical component \({B_v}\) of Earth’s magnetic field and so the magnetic flux linked with it changes. The potential difference or emf is induced. \(e = {B_v}lv\) Where \(l\) is the length of the axle, and \(v\) is the speed of the train.

The motion of an aeroplane in Earth’s magnetic field: A potential difference or emf across the wings of an aeroplane flying horizontally at a definite height is also induced because the aeroplane cuts the vertical component \({B_v}\) of Earth’s magnetic field. Thus, induced emf is \(e = {B_v}lv\) volt, the length \(l\) of an aeroplane’s wings, and the aeroplane’s speed \(v.\)

Orbital satellite: If the orbital plane of an artificial satellite of metallic surface coincides with the equatorial plane of the Earth, then no emf will be induced. If the orbital plane makes an angle with the equatorial plane, emf will be induced.

Motional EMF Due to Rotational Motion

1.  Conducting Rod

A conducting rod of length \(l\) whose one end is fixed is rotated about the axis passing through its fixed end and perpendicular to its length with constant angular velocity \(\omega .\) Magnetic field \((B)\) is perpendicular to the plane of the paper.

Emf induces across the ends of the rod \(e = \frac{1}{2}B{l^2}\omega = B{l^2}\pi v = \frac{{B{l^2}\pi }}{T}\)

where \(v=\) frequency (revolution per sec) and \(T=\) Time period.

Note : 

If the above metallic rod is rotated about its axis of rotation, then the induced potential difference between any pair of identical located points of the rod is always zero.

It is clear that parts \(O.P.\) and \(OQ\) are identical. 

Hence

\({e_{OP}} = {e_{OQ}}\) i.e. \({e_{OP}} = 0\)

Similarly \({e_{LN}} = 0\left( {{V_L} = {V_N}} \right)\)

 2. Cycle Wheel 

A conducting wheel each spoke of length rotates with angular velocity in a given magnetic field, as shown below in fig.

Due to flux cutting, each metal spoke becomes an identical cell of emf \(e\) (say), all such identical cells connected in a parallel fashion \({e_{net}} = e\) (emf of a single cell). Let \(N\) be the number of spokes hence \({e_{net}} = \frac{1}{2}B\omega {l^2};\omega = 2\pi v\)

Here \({e_{net}}\alpha {N^o},\) i.e. total emf does not depend on the number of spokes \(‘N’\)

Note: Here magnetic field (maybe a component of Earth’s magnetic field) sometimes depends on the plane of motion of the wheel. If the wheel rotates in a horizontal plane, \(B = {B_v};\) then used; If the wheel rotates in a vertical plane, then \(B = {B_H}({B_H} – \)horizontal component of Earth’s magnetic field while \({B_v} – \)vertical component)

 3. Faraday Copper Disc Generator 

During the rotational motion of the disc, it cuts away magnetic field lines.

A metal disc can be assumed to be made of uncountable radial conductors. When the metal disc rotates in the transverse magnetic field, these radial conductors cutaway magnetic field lines. Because of this flux cutting, all become identical cells each of emf \(‘e’\) where \(e = \frac{1}{2}B\omega {r^2},\)

As shown in following fig. and periphery of the disc becomes equipotential.

All identical cells are connected in parallel fashion, so net emf for disc

\({e_{net}} = e = \frac{1}{2}B\omega {r^2} = B\left( {\pi {r^2}} \right)v\)

Note: 

If a galvanometer is connected between two peripheral points or diametrical opposite ends, Its reading will be zero.

 4. Semicircular Conducting Loop 

For the given figure, a semicircular conducting loop \((ACD)\) of radius \(‘r’\) with centre at \(O,\) the loop plane is in the plane of the paper. The loop is now made to rotate with a constant angular velocity \(\omega ,\) about an axis passing through \(O\) and perpendicular to the plane of the paper. The effective resistance of the loop is \(R.\)

In time \(t\) the area swept by the loop in the field, i.e. region II    

\(A = \frac{1}{2}r\left( {r\theta } \right) = \frac{1}{2}{r^2}\omega t;\frac{{dA}}{{dt}} = \frac{{{r^2}\omega }}{2}\)

Flux link with the rotating loop at a time \(t,\)

\(\phi = BA\)

Hence induced emf in the loop in magnitude \(\left| e \right| = \frac{{d\phi }}{{dt}} = B\frac{{dA}}{{dt}} = \frac{{B\omega {r^2}}}{2}\) and induced current  \(i = \frac{{\left| e \right|}}{R} = \frac{{B\omega {r^2}}}{{2R}}\)

Periodic EMI

Suppose a rectangular coil of \(N\) turns is placed initially in a magnetic field such that the magnetic field is perpendicular to its plane, as shown-

\({\omega – }\)Angular speed 

\(V – \)Frequency of rotation of the coil 

\(R – \)Resistance of coil 

For uniform rotational motion with \(\omega ,\) the flux linked with the coil at any time

\(\phi = NBA\cos \theta = NBA\cos \omega t\left( {as{\mkern 1mu} \theta = \omega t} \right)\)

\(\phi = {\phi _0}\cos \omega t\) where \(\phi = NBA = {\rm{flux}}{\mkern 1mu} \,{\rm{amplitude}}\,{\mkern 1mu} {\rm{or}}\,{\mkern 1mu} {\rm{max}}\,{\mkern 1mu} {\rm{imum}}\,{\mkern 1mu} {\rm{flux}}\)

(This relation shows that the flux changes in periodic nature)

Induced EMF in the Coil

Induced emf also changes periodically. That’s why this phenomenon is called periodic EMI

\(e = – \frac{{d\phi }}{{dt}} = NBA\omega \sin \omega t \Rightarrow e = {e_0}\sin \omega t\)

Where \({e_0} = {\rm{emf}}\,{\rm{amplitude}}\,{\rm{or}}\,{\rm{max}}{\rm{.emf}} = NBA\omega = {\phi _0}\omega \)

Induced Current

At any time \(t,i = \frac{e}{R} = \frac{{{e_0}}}{R}\sin \omega t = {i_0}\sin \omega t\) where \({i_0} = \)current amplitude or max. current \({i_0} = \frac{{{e_0}}}{R} = \frac{{NBA\omega }}{R} = \frac{{{\phi _0}\omega }}{R}\)

Note: 

1. For the rotating coil, induced emf and linked flux keep a phase difference of \(\frac{\pi }{2},\) i.e. when the plane of the coil is perpendicular to the magnetic field \(\mathop B\limits^ \to ,\) so flux linked will be max. and induced emf  \(e = 0\) flux linked \({\phi _{\min }} = 0\)and induced emf will be maximum \({e_{\max }} = {e_0}\)
2. Frequency of induced any parameter\( = \)Frequency of rotation of coil\( = v\)
3. Both emf and current change their value w.r.t. time according to sine function; hence, they are called sinusoidal induced quantities.

Special Cases 

(i) A rectangular coil rotates at a constant speed about one of its sides A.B. The side A.B. is parallel to a long, straight current-carrying conductor.

The current-carrying conductor is in the plane of the page, and the magnetic field at the coil is perpendicular to the plane of the paper. The emf induced in the coil rotating in this field is minimum when the coil is perpendicular to the field in the plane of the conductor. The emf will be maximum when the coil is perpendicular to the plane of the conductor.

(ii) A stiff wire bent into a semicircle of radius \(‘r’\) is rotated at a frequency \(ν\) in a uniform field of magnetic induction \(\mathop B\limits^ \to \) as shown in the figure-

If the resistance of the entire circuit is \(R\) then

Current amplitude is given as \({i_0} = \frac{{BA\omega }}{R} = \frac{{B\left( {2\pi v} \right)}}{R}\frac{{\left( {\pi {r^2}} \right)}}{2} = \frac{{{\pi ^2}{r^2}Bv}}{R}\)

Area of loop \( = \frac{{\pi {r^2}}}{2}\)

(Frequency of induced current\(=\)frequency of rotation of loop\(=v\))

Solved Examples on Motional EMF

Q.1. A metal conductor of length \({\rm{1}}\,{\rm{m}}\) rotates vertically about one of its ends at an angular velocity of \(5\) radians per second. If the horizontal component of Earth’s magnetic field is \(0.2 \times {10^{ – 4}}T,\) then the e.m.f. developed between the two ends of the conductor is
(a) \(5mV\)
(b) \(50\mu V\)
(c) \(5\mu V\)
(d) \(50mV\)
Ans: (b) Induced emf \(e = \frac{1}{2}{B_H}{I^2}\omega = \frac{1}{2} \times 0.2 \times {10^{ – 4}} \times {\left( 1 \right)^2} \times 5 = 5 \times {10^{ – 5}}V = 50\mu V\)

Q.2. A copper disc of radius \({\rm{0}}{\rm{.1}}\,{\rm{m}}\) rotates about its centre with \(10\) revolutions per second in a uniform magnetic field of \(0.1\) Tesla. The emf induced across the radius of the disc is
(a) \(\frac{\pi }{{10}}V\)
(b) \(\frac{{2\pi }}{{10}}V\)
(c) \(10\,\pi \,mV\)
(d) \(20\,\pi \,mV\)
Ans: (c) The induced emf between the centre and rim of the rotating disc is \(E = \frac{1}{2}B\omega {R^2} = \frac{1}{2} \times 0.1 \times 2\pi \times 10 \times {\left( {0.1} \right)^2} = 10\pi \times {10^{ – 3}}{\rm{volt}}\)

Q.3. A two-metre wire moves with a velocity of \({\rm{1m/sec}}\) perpendicular to a magnetic field of \({\rm{0}}{\rm{.5}}\,{\rm{weber/}}{{\rm{m}}^2}.\) The emf induced in it will be
(a) \({\rm{0}}{\rm{.5}}\,{\rm{volt}}\)
(b) \({\rm{0}}{\rm{.1}}\,{\rm{volt}}\)
(c) \({\rm{1}}\,{\rm{volt}}\)
(d) \({\rm{2}}\,{\rm{volt}}\)
Ans: (c) \({\rm{e = BVl = 0}}{\rm{.5}} \times {\rm{1}} \times {\rm{2 = 1}}\,{\rm{volt}}\)

Q.4. The two rails of a railway track, insulated from each other and the ground, are connected to a millivoltmeter. What is the reading of the millivoltmeter when a train travels at a speed of \({\rm{20}}\,{\rm{m/sec}}\) along the track, given that the vertical component of Earth’s magnetic field is \(0.2 \times {10^{ – 4}}\,{\rm{wb/}}{{\rm{m}}^{\rm{2}}}\) and the rails are separated by \({\rm{1}}\,{\rm{metre}}\)
(a) \(4\,mV\)
(b) \(0.4\,mV\)
(c) \(80\,mV\)
(d) \(10\,mV\)
Ans: (b) When a train runs on the rails, it cuts the magnetic flux lines of the vertical component of Earth’s magnetic field. Hence a potential difference is induced between the ends of its axle. Distance between the rails \({\rm{l = 1}}\,{\rm{m}}{\rm{.}}\)
Speed of train \(v = 36\frac{{{\rm{km}}}}{{{\rm{hour}}}} = \frac{{36 \times 1000}}{{3600}} = {\rm{10}}\,{\rm{m/sec}}\)
By using   \(e = Bvl{\rm{ }};{\rm{ }}e = 0.2 \times {10^{–4}} \times 20 \times 1 = 4 \times {10^{–4}}{\rm{volt}} = 0.4{\rm{ }}mV\)

Q.5. As shown in the figure, a metal rod makes contact and complete the circuit. The circuit is perpendicular to the magnetic field with \({\rm{B = 0}}{\rm{.15}}\) Tesla. If the resistance is \({\rm{3}}\Omega {\rm{,}}\) the force needed to move the rod as indicated with a constant speed of \({\rm{2}}\,{\rm{m/sec}}\)is

(a) \(3.75 \times {10^{ – 3}}N\)
(b) \(3.75 \times {10^{ – 2}}N\)
(c) \(3.75 \times {10^{ – 2}}N\)
(d) \(3.75 \times {10^{ – 4}}N\)
Ans: (a) Force needed to move the rod is \(F = \frac{{{B^2}v{l^2}}}{R} = \frac{{{{\left( {0.15} \right)}^2} \times 2 \times {{\left( {0.5} \right)}^2}}}{3} = 3.75 \times {10^{ – 3}}N\)

Summary

1. A piece of metal and a piece of non-metal are dropped from the same height near the surface of the Earth. The non-metallic piece will reach the ground first because there will be no induced current in it.
2. If an aeroplane is landing down or taking off and its wings are in the east-west direction, then the potential difference or emf will be induced across the wings. If an aeroplane is landing down or taking off and its wings are in the north-south direction, no potential difference or emf will be induced.
3. When a conducting rod moves horizontally on Earth’s equator, no emf induces because there is no vertical component of Earth’s magnetic field. But at poles, is maximum, so maximum flux cutting takes place. Hence emf induces.
4. When a conducting rod is fallingly in Earth’s magnetic field, its length lies along East-West direction then induced emf continuously increases w.r.t. time and induced current flows from West – East.

FAQ’s on Motional EMF

Q.1. Two spherical bobs, one metallic and the other of Glass, of the same size, are allowed to fallly from the same height above the ground. Which of the two would reach earlier and why?
Ans: The glass bob will reach earlier on the ground as acceleration due to gravity is independent of the mass of the falling bodies. Glass is an insulator. No induced current is developed in it due to the Earth’s magnetic field.

Q.2. What is motional emf? Explain.
Ans: An emf induced by motion relative to a magnetic field is called a motional emf. The equation which represents this emf is \(e = BVl,\) where \(l\) is the object’s length moving at a speed \(v\) relative to the strength of the magnetic field \(B.\)

Q.3. How can we generate a motional emf?
Ans: Motional emf can be induced in two major ways:
I. Due to the motion of a conductor in the presence of a magnetic field.
II. Due to the change in the magnetic flux enclosed by the circuit.

Q.4. What are the factors on which the motional emf depends?
Ans: The factors on which Motional emf depends are the magnetic field and velocity, and length of the rod.

Q.5. What is the S.I. unit of motional emf?
Ans: The S.I. unit of motional emf is volt or joule/coulomb.