CHAPTER-29

1. CHAPTER-29 Magnetic Fields Due to Currents

2. Ch 29-2 Calculating Magnetic Field Due to a Current • Biot Savart Law • Magnetic field dB due to a differential current-length element ids-element of a length vector ds in the direction of current i at a point P at a distance r from the current-length element ds is given by: • dB (i ds sin)/r2 • dB=(0/4) (i ds sin)/r2 • dB=(0/4) (i ds x r)/r3(Biot-Savart)

3. Ch 29-2 Calculating Magnetic Field Due to a Current • Magnetic Field lines due to a Current in a Long Straight Wire encircles the wires clockwise ( current into the page) or counter clockwise ( current pout of the page). • Direction of B is tangent to those circles in the direction of field lines

4. Ch 29-2 Calculating Magnetic Field Due to a Current • Magnetic Field Due to a Current in a long straight wire • Right hand rule: Grasp the element in your right hand with your extended thumb pointing in the direction of the current. Your finger will then naturally curl around in the direction of magnetic field line due to that element.

5. Ch 29-2 Calculating Magnetic Field Due to a Current • Magnetic Field Due to a Current in a long straight wire B= 0i/2R • Magnetic Field Due to a Current in a semi-long straight wire B= 0i/4R • Magnetic Field Due to a Current in a Circular Arc of Wire B= (0i/4R) (rad)

6. Ch 29-3 Forces Between Two Parallel Currents • Force on Wire b of length L with current ib due to magnetic field of very long wire a at the location of b • Fba=ib Lx Ba and Ba= 0ia/2d • Fba=ib Lx Ba=0iaib (Lx Ba )/2d • Fba= 0iaib (LBa sin)/2d • Fba= 0iaib LBa /2d • Fba/L= 0iaib LBa /2d • Parallel currents attracts each other • Antiparallel currents repels eaxh other

7. Ch 29-4 Ampere’s Law • Ampere’s Law: magnetic analogue of Gauss’s law • Gauss’ law: surf E.dA= qenc/0 • Ampere’s Law: loop B.ds= 0 ienc integral to be evaluated along Ampereian loop, ienc is net current enclosed by the loop • Curled-straight right-hand rule to determine sign for current direction: Curl your right hand rule around the Amperian loop, with the fingers pointing in the direction of integration. A current through the loop in the direction of outstretched thumb is +ve and opposite is –ve. • loop B.ds= loop B ds cos=0 ienc=0 (i1-i2) •  is angle between B and ds • loop Bds cos=loop Bds= Bloop ds=0 (i1-i2) B2r=0 ienc

8. Ch 29-4 Ampere’s Law • Magnetic Field outside a long straight wire with a current: • B= loop Bds cos • = Bloop ds=B2r=ienc (=0) B2r=0 ienc (outside a very long wire) • Magnetic Field inside a long straight wire with a current: • B= 0 i’enc/2r i’enc= (ienc/R2)r2= iencr2/R2 • B= 0 i’enc/2r= (0 iencr)/2R2

9. Ch 29-5 Solenoids and Toroids Solenoids: Helical coil of wire producing uniform magnetic field at the center. • Length of the coil is much greater than the coil radius • B-field of the solenoid: loop B.ds= 0 ienc loop B.ds= abB.ds + bcB.ds+ cdB.ds+ daB.ds= = abB.ds= Bh ienc=iN=inh Then Bh= 0 inh • B=0 in n is number of turns per unit length of the solenoid.

10. Ch 29-5 Solenoids and Toroids • Toroid: Hollow solenoid curved to meet its two ends. • B field inside a torus is given by: • loop B.ds= 0 ienc • loop B.ds= B 2r=0 Ni B=0 Ni/2r

11. Ch 29-6 A Current Carrying Coil as a Magnetic Dipole Z component of magnetic field of a current carrying coil Bz given by: BBz =(0 /2)(NiA) /z3 =(0 /2) /z3 where z-axis is from south to north axis

12. Suggested problems Chapter 29