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Biot-Savart law

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Use the Biot-Savart law for a straight segment of wire, to obtain the flux density at the center of a rectangular coil of dimensions 5cm x 3cm, wound with 100 closely-spaced turns carrying 0.25 A.

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Above fig. shows a straight current carrying conductor. Point P is at a perpendicular distance a from the conductor. Angles Î¸1 and Î¸2 are as shown in the fig.. Magnetic flux density B at P due to the current carrying conductor is given by: B = (Î¼0I/4ÐŸa)[sinÎ¸1 + sinÎ¸2] (both angles to be taken as +ve) ..........(1)

Direction of vector B for the given direction of current is into the plane (reverse direction of current it will be out of the plane).

Magnetic flux density at the centre of a ...

Solution Summary

Using the Biot-Savart law for a straight segment of wire are determined. Step by step solution provided.

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