Question

A horizontal conducting rod $10m$ long extending from east to west is falling at a speed of $5.0m/s$ at right angles to the horizontal component of earth’s magnetic field $0.3\times {{10}^{-4}}Wb/{{m}^{2}}$ .Find the instantaneous values of e.m.f induced in the rod.

Answer 1

Hint The magnetic force is encountered by charges in the loop's wires when they shift in a magnetic field. Charges in the vertical wires undergo forces that generate currents parallel to the cable. However, a force perpendicular to the wire is felt by those in the top and bottom segments; this force does not create a current.

Complete step by step answer
In the above question following data is been given –
           $v=5.0m/s$
            $l=10m$
           $B=0.3\times {{10}^{-4}}Wb/{{m}^{2}}$
Where, v denotes the velocity with which the rod is falling.
               L denotes the length of the rod.
              B denotes the magnetic field that is being induced.
We all know that because of its complex relationship with the magnetic field, EMF is induced in it when an electrical conductor is inserted into a magnetic field. This EMF is known as EMF mediated.
This induced emf is referred to as motional emf due to the motion of an electrical conductor in the presence of the magnetic field. EMF can thus be caused in two primary ways:
In the existence of a magnetic field regardless of the action of a conductor.
Because of the magnetic flux shift that is surrounded by the circuit.
With the help of the concept of Lorentz force acting on the conductor's free charge carriers, this definition of motional emf can be clarified. Let us assume, in the conductor PQ, any random charge q. As the rod travels at steady velocity v, in the presence of magnetic field B, the charge still travels at velocity v. The power of Lorentz on this charge is given by:$\Rightarrow e=1.5\times {{10}^{-3}}volt$
$F=qvB$
And now let us have a look at the total work done
$W=QBvl$
Now, we know that any type of electromagnetic force induced is work done per unit charge, therefore we have,
$e=wq=Bvl$
In the above question all the data is already given we just have to put the values in the above formula-
$e=0.3\times {{10}^{-4}}\times 0.5\times 10$
$\Rightarrow e=1.5\times {{10}^{-3}}volt$
Thus we have our answer.

Note The Lorentz force 's direction is perpendicular to both the current flow direction and the magnetic field and can be found using the right-hand law seen in it. Point your thumb in the direction of the current using your right palm, and point the first finger in the direction of the magnetic field. Now your third finger is going to point in the direction of the force.