CHAPTER-29

Magnetic Fields Due to Currents

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)

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

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.

Ch 29-2 Calculating Magnetic Field Due to a Current

Magnetic Field Due to a Current in a long straight wire B= 0i/2R Magnetic Field Due to a Current in a semi-long straight wire B= 0i/4R Magnetic Field Due to a Current in a Circular Arc of Wire B= (0i/4R) (rad)

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/2d

Fba=ib Lx Ba=0iaib (Lx Ba )/2d

Fba= 0iaib (LBa sin)/2d

Fba= 0iaib LBa /2d

Fba/L= 0iaib LBa /2d

Parallel currents attracts each other

Antiparallel currents repels eaxh other

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= Bloop ds=0 (i1-i2)

B2r=0 ienc

Ch 29-4 Ampere’s Law

Magnetic Field outside a long straight wire with a current:

B= loop Bds cos = Bloop ds=B2r= ienc (=0)

B2r=0 ienc (outside a very long wire)

Magnetic Field inside a long straight wire with a current:

B= 0 i’enc/2r

i’enc= (ienc/R2)r2= iencr2 /R2

B= 0 i’enc/2r= (0 iencr)/2R2

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=

ab

B.ds + bc B.ds+ c

d B.ds+ d

a

B.ds=

= ab

B.ds= Bh

ienc=iN=inh

Then Bh= 0 inh

B=0 in

n is number of turns per unit length of the solenoid.

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 2r=0 Ni

B=0 Ni/2r

Ch 29-6 A Current Carrying Coil as a Magnetic Dipole

Z component of magnetic field of a current carrying coil Bz given by:

BBz =(0 /2) (NiA) /z3

=(0 /2) /z3

where z-axis is from south to north axis

Suggested problems Chapter 29