Suppose we have a closed loop with two resistors in series and at particular moment in time the induced emf was two volts .i want to know why do we use ohm's law in calculating the induced current ?
Current isn’t induced; voltage is induced as per Faraday’s law of induction and, as a result of voltage being induced (by a changing magnetic field), if there is a current flow then that is defined by the induced voltage and the impedance of the circuit i.e. Ohm’s law.
At DC use resistance and for an AC induced voltage use the reactance and resistance and remember to consider that the wire in which the voltage is induced has self-inductance.
If you want to induce a voltage in a loop, that loop having one side parallel to a long wire, and the wire and the loop are in the same plane, then you can use this formula
Vinduce = [MUo * MUr * Area / (2 * PI * Distance)] * dI/dT
where we avoid using natural-logs in the equation by assuming the separation of the wire and the loop is large compared to the sides of the loop. If these assumpions are wrong, you'll have a 2;1 or 3:1 error, but you'll still get a good idea of the size of the induced voltage, which lets you ponder the size of any magnetic-field interference.
How to use this? assume the dI/dT is 2,000 amps per microsecond (the current is being switched by IGBTs in a massive motor speed controller, perhaps for a ore-grinding mill)
Let the distance be 4 centimeters. Let the area be of the GROUND plane size of the SpeedController PCB board.
What are the numbers? With MUo of 4 * pi * 1.0e-7, and MUr = 1 (air, copper), we have
Vinduce = 2e-7 * Area/Distance * dI/dT
Vinduce = 2e-7 * 0.1meter * 0.1meter / 0.04meters * 1e+9 amp/sec
Vinduce = 2e-7 * 0.010 / 0.04 * 1e+9
Vinduce = 2 * 0.01/0.04 * e-7 * e+9= 0.5 * 100 = 50 volts
Thus, before any induced eddy currents that oppose and almost null the incoming magnetic field, there is 50 volts imposed on the sheet of copper, the Ground plane.
Given 0.000500 ohms (500 microOhms) per square of standard thickness PCB copper foil, the eddy current easily reaches the level of 50/0.0005 = 50 * 2,000 = 100,000 amps to cancel the incoming Hfield.