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This is a good question. The answer depends on what you connect to the two leads of the coil. If you let it open, there will be no current induced in the coil, therefore no effect on the magnetic field. If you short-circuit the two leads, and the resistance of the coil is very small, there will be a very strong induced magnetic field which will almost cancel out the imposed magnetic field. Consider the magnetic field penetrating the coil. It is composed of two parts: one is the external magnetic field B_e which does not depend on the current in the coil, another is the internal magnetic field B_i caused by the current I in the coil. A useful quantity related to this problem is the magnetic flux. If you view magnetic field as a flowing stream, the magnetic flux is the quantity of magnetic field flowing through the coil. The magnetic flux F_i due to B_i is proportional to the current I in the coil, F_i = I * L Actually L is just the inductance of the coil. Denote the magnetic flux due to B_e as F_e, there will be a voltage induced at the two leads of the coil, which is d dI dF_e V = -- (F_i + F_e) = L * -- + ----- dt dt dt Therefore, if you draw a equivalent circuit for the coil, it is a inductance L in series with a voltage source Vs = dF_e/dt. For a current to appear in the coil, you need to complete the circuit by connecting the two leads of the coil with a wire or a resistor or a capacitor or anything which can let a current pass through it. The the current will be determined by the whole circuit. For example, if the external magnetic field is a sinusoidal function of time with angular frequency as w, then you will get a sinusoidal voltage source Vs = jw*F_e (Here for simplicity we use phasor to represent the quantities and assume the time dependence as exp(j*w*t), j is the square root of -1). Assume you connect a resistor R at the two leads of your coil, then the current in the coil will be -Vs -jw*F_e I = ----- = -------- jwL+R jwL+R The total magnetic flux through the coil is F_i + F_e = L * I + F_e = R * F_e / (jwL + R) Therefore, if you let the coil open-circuited, that is, you connect nothing to its two leads, the magnetic field through the coil will not be affected by the coil, since there is no current induced in it. However, If you connect the two leads of the coil through a wire, and the coil and the wire are very good conductance, the magnetic field caused by the induced current will be so strong that it will almost totally cancel out the imposed external magnetic field. Some tutorials on phasors: http://www.clarkson.edu/~svoboda/eta/phasors/Phasor10.html http://www.clarkson.edu/~svoboda/eta/phasors/MatchPhasors10.html http://www.clarkson.edu/~svoboda/eta/phasors/AddPhasors10.html

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