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Welcome to our Physics lesson on Induced Electromotive Force as Motional Emf, this is the third lesson of our suite of physics lessons covering the topic of Faraday's Law of Induction, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Let's extend the discussion about the induced emf considering it from another viewpoint. As a special case of electromagnetic induction is when a magnetic force acts on a straight current carrying wire placed inside a uniform magnetic field is moving due right as shown in the figure below.
The direction of the resulting magnetic force is vertically down, based on the Fleming's Left Hand Rule explained in the previous tutorials. The scalar equation of this magnetic force is
Since the wire is moving due right, the magnetic force is balanced by another force acting in the upward direction. This force is the electric force Fe produced on the moving charges.
We can write the equation that shows mathematically this balance of forces as
Since the electric field produced by the moving charges is constant, we express the relationship between the electric field E and the potential difference ΔV across the ends of conducting wire as
Combining the last two equation, we obtain
This potential difference is equal to the emf ε of the wire as the resistance of wire is negligible. Obviously, this is an induced emf as it is caused due to the motion of wire and is not constant; it depends on the following factors:
Hence, the formula of emf induced in the conducting wire when it moves inside a magnetic field is
If the conductor that is sliding due right is part of a closed conducting path as shown in the figure below,
then, electrons will move in the clockwise direction throughout the circuit due to the existence of emf produced by the source. (Remember the direction of conventional current is opposite to the direction of electron's flow).
The value of current I flowing through the circuit in this case, is
where R is the resistance in the circuit.
At what speed should we move a 50 cm bar as shown in the figure below to produce a current of 0.4 A if the bar touches the two opposite sides of a circuit having a resistance of 20 Ω? The plane of circuit (including the bar) is normal to a 2 T magnetic field the direction of which is into the page.
Clues:
L = 50 cm = 0.50 m
I = 0.4 A
R = 20 Ω
B = 2 T
v = ?
From the equation
we obtain for the speed v after rearranging the terms
You have reached the end of Physics lesson 16.7.3 Induced Electromotive Force as Motional Emf. There are 4 lessons in this physics tutorial covering Faraday's Law of Induction, you can access all the lessons from this tutorial below.
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