SHIELDING A RECEPTOR AGAINST MAGNETIC FIELDS
The best way to protect against magnetic fields at the receptor is to decrease the area of the receptor loop. The area of interest is the total area enclosed by current flow in the receptor circuit. An important consideration is the path taken by the current in returning to the source. Often, the current returns by a path other than the one intended by the designer, and therefore, the area of the loop changes. If a nonmagnetic shield placed around a conductor causes the current to return over a path that encloses a smaller area, then some protection against magnetic fields will have been provided by the shield. This protection, however, is caused by the reduced loop area and not by any magnetic shielding properties of the shield.
Figure below illustrates the effect of a shield on the loop area of a circuit. In Fig.A, the source VS is connected to the load RL by a single conductor, using a ground return path. The area enclosed by the current is the rectangle between the conductor and the ground plane. In Fig.B, a shield is placed around the conductor and grounded at both ends. If the current returns through the shield rather than the ground plane, then the area of the loop is decreased, and a degree of magnetic protection is provided. The current will return through the shield if the frequency is greater than five times the shield cutoff frequency as previously shown. A shield placed around the conductor and grounded at one end only, as shown in Fig. C, does not change the loop area and therefore provides no magnetic protection.
The arrangement of Fig. B does not protect against magnetic fields at frequencies below the shield cutoff frequency because then most of the current returns through the ground plane and not through the shield. At low frequencies, this circuit also has two other problems, as follows: (1) Because the shield is one of the circuit conductors, any noise current in it will produce an IR drop in the shield and appear to the circuit as a noise voltage, and (2) if there is a difference in ground potential between the two ends of the shield then it will show up as a noise voltage in the circuit.