the electric current in the direction B TO A is decreasing. what is the direction of induced current in the loop kept above the wire.
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Due to the current I1 through the wire B to A, there is a magnetic flux Φ1 through the loop above the wire BA. It (Φ1) is equal to integral of B. dA in the area enclosed by the loop. B = μ I1 / (2 π d). Since, flux is proportional to the current I1 in the wire, it decreases when current I1 decreases.
We have to use the Faraday's law and Lenz's law for finding the emf generated and for the direction of the current. The direction is given by right/left hand rule or left/right hand thumb rule.
induced emf in the loop = emf = - dΦ1 / dt
induced current = I2 = emf / R , where R = resistance in the loop
The direction of the current in the loop will be such that it produces magnetic flux Φ2 through the loop, in the direction opposite to ΔΦ1. ΔΦ1 is negative, as the flux Ф1 is decreasing through the loop due to decreasing current I1. Hence, I2 will flow to increase flux through the loop.
Applying the conservation of energy principle, the current flowing with the energy that the magnetic field gives it and that energy comes from the work done by the agent that reduces current in the wire AB.
See the diagram to understand this.
We have to use the Faraday's law and Lenz's law for finding the emf generated and for the direction of the current. The direction is given by right/left hand rule or left/right hand thumb rule.
induced emf in the loop = emf = - dΦ1 / dt
induced current = I2 = emf / R , where R = resistance in the loop
The direction of the current in the loop will be such that it produces magnetic flux Φ2 through the loop, in the direction opposite to ΔΦ1. ΔΦ1 is negative, as the flux Ф1 is decreasing through the loop due to decreasing current I1. Hence, I2 will flow to increase flux through the loop.
Applying the conservation of energy principle, the current flowing with the energy that the magnetic field gives it and that energy comes from the work done by the agent that reduces current in the wire AB.
See the diagram to understand this.
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