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AMPERE’S LAW

AMPERE’S LAW. SLIDES BY ZIL E HUMA. OBJECTIVES. THE BIOT-SAVART LAW AMPERE’S CIRCUIT LAW. THE BIOT-SAVART LAW. We know develop a procedure for calculating the magnetic field due to a specify current distribution. Before considering the magnetic field.

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AMPERE’S LAW

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  1. AMPERE’S LAW SLIDES BY ZIL E HUMA

  2. OBJECTIVES • THE BIOT-SAVART LAW • AMPERE’S CIRCUIT LAW

  3. THE BIOT-SAVART LAW • We know develop a procedure for calculating the magnetic field due to a specify current distribution. Before considering the magnetic field. Let us first review the procedure for calculating the magnetic field.

  4. q1 dq1 q2 dq2 ur dE1

  5. Figures shows the two charge distribution q1 and q2 of arbitrary size and shape. We consider charge elements dq1 and dq2 in the two distributions. The electric field dE1 set up by dq1 at the location of dq2 is given by dE1 = 1/40 dq1/r2 (ur) = (1/ 40 ) (dq1 / r3 ) r where r is the vector from dq1 to dq2 and r is its magnitude,

  6. ur is the unit vector in the direction of r . The electric field dE1 set up by dq1 at the location of dq2 is given by dE1 = 1/40 dq1/r2 (ur) = (1/ 40 ) (dq1 / r3 ) r --------- 1 where r is the vector from dq1 to dq2 r is its magnitude, ur is a unit vector in the direction of r.

  7. To find the total electric field E1 acting at dq2 due to the entire distribution q1, we integrate over q1: E1 = 1/40dq1/r2 (ur) = (1/ 40 )(dq1 / r3 ) r ----------- 2 The force dF21 acting on dq2 due to the charge distribution q1 can then be written as dF21 = E1 dq2 --------------- 3

  8. Eq. 1 or 2 gives the electric field of the charge distribution and eq. 3 giving the force due to that distribution acting on another charge. The all three equations together can be taken, as the column’s law for finding the electrostatic force between charges .

  9. In the case of magnetic fields we see the force between current elements i1 dB1 ds1 r ds2 ur i2 Two arbitrary current distributions i1 and i2

  10. We consider to current i1 and i2 and their corresponding current elements i1ds1 and i2ds2 the magnetic force dF21 exerted on current element 2 by i1 can be written dF21 = i2ds2 * B1 -------- 4 Where the magnetic field B1 at the location of current element i2ds2 is due to entire current i1

  11. The contribution dB1 of each current element of i1 to the total field B1 is given by dB1 = k(i1ds1 * ur/r2) = k (i1ds2 *r/r3)-------------5 Where r is the vector from current element 1 to current element 2 and the ur is the unit vector in the direction of r Eq. No. 4 & 5 together give the magnetic force between the current elements.

  12. CONSTANT k On determine constant k, here we included a similar constant, as in coulomb's law. For determining the constant in the magnetic force law. Set the constant equal to a convenient value and use the force law to define the unit of current, the ampere.

  13. The constant k in SI units is defined to have the exact value 10-7 tasla.meter/ampere. In electrostatics, we find it convenient to write the constant in a different form: k =0/4 = 10-7 T.m/A, where the constant 0 called the permeability constant has the exact value 0 = 4 * 10-7 T.m/A,

  14. Dropping the subscripts in Eq. No. 5 and using the eq. No. 6 for the constant k we have. dB1 = 0/4 (i1ds1 * ur/r2) = 0/4 (i1ds2 *r/r3) --------7 This result is known as the biotsavrt law.The direction of the db is the same as the direction of ds * ur into the plane of the paper. Magnitude of the dB1 = 0/4(i1ds1 sin/r2) 8 Where is the angle ds and r.

  15. To find the total field b due to the entire current distribution, we must integrate over all current elements ids. B =  dB =0/4 ids * ur/r2 =0/4 ids * r/r3 --------- 9

  16. AMPERE’S CIRCUIT LAW : We know that an electric current produces a magnetic field. Ampere performed a number of experiments and he showed that “the flux density B at any point due to a long straight conductor was directly proportional to the strength of the current and inversely proportional to the distance of the point from the conductor.”

  17. i r P

  18. If the current flowing through a conductor is i, the flux density at a point distance ‘r’ from it is given by B (i/r) Or B = (0/2) (i/r).

  19. Where0/2is the constant of proportionality. 0 is called the permeability of free space and its value is 4 * 10-7weber/ amp-m.

  20. Again Ampere considered a long conductor carrying a current i with a circular line of magnetic force of radius ‘r’ as show in figure. The flux density at every point of the circle of radius r is given by B= (0 /2)(i/r) Or 2r * B=0 i.

  21. 2r is clearly the total path around the conductor over which the B has the same value at every point. Ampere generalized this result into a law, which states that the flux density or the magnetic field of a closed path due to a current is always equal to the product of the permeability and the current I.e. 0 i . This law is know as Ampere’s circuit law.

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