
Predict the direction of induced current in a metal ring when the ring is moved towards a straight conductor vertically down with constant speed. The conductor is carrying current in the East direction.
Answer
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Hint: We know that according to Lenz law the induced current will always be in the direction in which it opposes the change of flux in conductor. There is some magnetic field due to current carrying wire and due to its motion of conductor the flux induced in conductor is changed so, there is a change of flux induced in conductor ring so, there is an induced current so, the flux generated by this current opposes the change of flux change.
Complete solution:
We know that a current carrying conductor generates a magnetic field nearby. Its value is given by expression in the case of straight current carrying wire is
$B=\dfrac{{{\mu }_{0}}I}{2\pi r}$
Where, $B$ is the intensity of magnetic field generated by current carrying wire, ${{\mu }_{0}}$ is the permeability of free space, $I$ is the current flowing in the wire, $r$ is the shortest distance between the wire and that point where we are calculating the intensity of magnetic field.
So, we can see that decreasing the distance from the wire results in an increase in intensity of the magnetic field. Also by right hand rule the direction of intensity can be determined that is perpendicular upward to the plane of the ring conductor or plane of paper. So, there is change in flux induced in this conductor it results in generation induced Emf (electromotive force) in moving conductor so, due to this emf a current is flow in such a way by which the magnetic field will be in downward direction so, by right hand thumb rule for downward direction of magnetic field the direction of current will in clockwise direction.
Note: We have to learn that the right hand thumb rule is used for determining the direction of magnetic field intensity due to a current carrying conductor, while we determine the direction of force experience by the conductor on a magnetic field by Fleming’s left hand rule. Also there are different formulae for magnetic field intensity for different shapes of current carrying conductors
Complete solution:
We know that a current carrying conductor generates a magnetic field nearby. Its value is given by expression in the case of straight current carrying wire is
$B=\dfrac{{{\mu }_{0}}I}{2\pi r}$
Where, $B$ is the intensity of magnetic field generated by current carrying wire, ${{\mu }_{0}}$ is the permeability of free space, $I$ is the current flowing in the wire, $r$ is the shortest distance between the wire and that point where we are calculating the intensity of magnetic field.
So, we can see that decreasing the distance from the wire results in an increase in intensity of the magnetic field. Also by right hand rule the direction of intensity can be determined that is perpendicular upward to the plane of the ring conductor or plane of paper. So, there is change in flux induced in this conductor it results in generation induced Emf (electromotive force) in moving conductor so, due to this emf a current is flow in such a way by which the magnetic field will be in downward direction so, by right hand thumb rule for downward direction of magnetic field the direction of current will in clockwise direction.
Note: We have to learn that the right hand thumb rule is used for determining the direction of magnetic field intensity due to a current carrying conductor, while we determine the direction of force experience by the conductor on a magnetic field by Fleming’s left hand rule. Also there are different formulae for magnetic field intensity for different shapes of current carrying conductors
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