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CHAPTER 30 : SOURCES OF THE MAGNETIC FIELD 30.1) THE BIOT-SAVART LAW To calculate the magnetic field produced at some point in space by a small current element.
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These observations are summarized in the mathematical formula known today as the Biot-Savart Law :
where the integral is taken over the entire current distribution.
If the conductor lies in the plane of the page (Figure (30.1)), dB points out of the page at P and into the page at P’ (Difference (1)).
Example (30.2) : Magnetic Field Due to a Curved Wire Segment
Calculate the magnetic field at point O for the current-carrying wire segmetn shown in Figure (30.4). The wire consists of two straight portions and a circular are of radius R, which subtends an angle . The arrowheads on the wire indicate the direction of the current.
where (the circumference of the loop)
(at x = 0)
(for x >> R
The direction of F1 is toward wire 2 because is in that direction.
When the magnitude of the force per unit length between two long, parallel wires that carry identical currents and are separated by 1m is 2 x 10-7N/m, the current in each wire is defined to be 1A.
The value 2 x 10-7N/m is obtained from Equation (30.12) with I1 = I2 = 1A, and a = 1m.
When a conductor carries a steady current of 1A, the quantity of charge that flows through a cross-section of the conductor in 1s is 1C.
Let us evaluate the product B·ds for a small length element ds on the circular path defined by the compass needles, and sum the products for all elements over the closed circular path.
The line integal of B·ds around any closed path equals oI, where I is the total continuous current passing through any surface bounded by the closed path.
(for r < R) (30.15)
Example (30.6) : Magnetic Field Created By an Infinite Current Sheet
To evaluate the line interal in Ampere’s law, let us take a rectangular path through the sheet, (Figure (30.14)).the rectangel has dimensions parallel to the sheet surface.
The field lines between current elements on two adjacent turns tend to cancel each other because the field vectors from the two elements are in opposite directions.
Because the solenoid is ideal, B in the interior space is uniform and parallel to the axis, and B in the exterior space is zero.
In this case, the total current through the rectangular path equals the current through each turn multiplied by the number of turns.
Magnetic field inside a solenoid
Definition of Magnetic Flux
Example (30.8): Magnetic Flux Through a Rectangular Loop
A rectangular loop of width a and length b is located near a long wire carrying a current I (Figure (30.21)). The distance between the wire and the closest side of the loop is c. the wire is parallel to the long side of the loop. Find the total magnetic flux through the loop due to the current in the wire.
(Because B is not uniform but depends on r, it cannot be removed from the integral)
Gauss’s Law for magnetisim