Chapter 3 Magnetostatic ECT1026 Field Theory Chapter 3 Magnetostatic Lecture 3-3 Application of Biot-Savart Law By Dr Mardeni Roslee mardeni.roslee@mmu.edu.my 0383125481

Application of Biot-Savart Law ECT1026 Field Theory Lecture 3-3 Application of Biot-Savart Law Magnetic Field of a Pie-Shaped Loop Circular Loop Linear Conductor

B ? Example 3.2-1: Magnetic Field of a Linear Conductor ECT1026 Field Theory Example 3.2-1: Magnetic Field of a Linear Conductor Using Biot-Savart law to find the magnetic flux density B at a point in the plane that bisects a long straight wire (or point P) of length 2L carrying a direct current I. P L B ?

ECT1026 Field Theory How to solve ? Step-1: Cylindrical coordinate system - Imagine the wire lies along the z-axis Step-2: Let the bisecting plane be the xy-plane z ˆ (r,,z) I ˆ x y

ECT1026 Field Theory How to solve ? Step-1: Cylindrical coordinate system - Imagine the wire lies along the z-axis Step-2: Let the bisecting plane be the xy-plane z ˆ z ˆ P r ^ L - L r = r r Simplified I r ˆ x y r ^  P (r,,z)

ECT1026 Field Theory How to solve ? Step-3: Start by considering the B generated by a current element dl= z dz at a distance R from the current element ^ z ˆ P r ^ L - L r = r r I R = - z r ˆ Therefore = z dz (r r – z z ) ˆ dl  R  ^ dz z R r  z 0 0 dz r 0 z ^ =  r dz dl  R =  r dz ^   dl  R = ^  r dz R R = R R ^

I dB = mo dl × R R 4 4p ˆ dB = mo I 4 R rdz dB = rdz mo I  ECT1026 Field Theory How to solve ? R 2 mo dB = 4p dl × R I ^ Step-4: Biot-Savart Law z ˆ P r ^ L - L r = r r I dl  R = ^  r dz R Since dz z R r z R dB = mo I 4 R 3 rdz ^ dB = (z2 + r2) 3/2 rdz mo I 4  ^ R2 = z2 + r2

∫ ∫ B = dB ˆ moIr  B = 4 mo Ir  B = 4 mo IL  B = How to solve ? ECT1026 Field Theory How to solve ? +L B = ∫ dB Step-5: Integration -L z ˆ P r ^ L - L r = r r I  ^ B = (z2 + r2) 3/2 dz ∫ -L +L moIr 4 dz z R  ^ B = mo Ir 4 r2 (z2 + r2) 1/2 z -L +L  ^ B = r (L2 + r2) 1/2 1 mo IL 2

For an infinitely long wire such that L >> r ˆ  ^ B = r (L2 + r2) 1/2 1 mo IL 2 For an infinitely long wire such that L >> r z ˆ P r ^ L - L r = r r I  ^ B = mo I 2r  ^ B = (L2 + r2) 1/2 L mo I 2r L >> r,  ^ B = (L2 + 02) 1/2 L mo I 2r

Divide the circular loop into two segments: ECT1026 Field Theory Example 3.2-2: Magnetic Field on a Pie-Shaped Loop A circular loop of radius a carries a steady current I. Determine the magnetic field B at a the origin O of the loop. Divide the circular loop into two segments: (ii) AC (i) OA & OC B ?

dl is parallel or anti-parallel to R hence dl  R = 0 dl ECT1026 Field Theory Step 1: Segment OA and OC For the straight segments OA and OC, the magnetic field at point O is zero. As all points along these segments, dl is parallel or anti-parallel to R hence dl  R = 0 O C A ^ dl ^ R ^ Parallel R ^ Anti-parallel AB = AB sin  sin 0 = sin 180 = 0

dl =a d A C dl =  a d r = - R a or r = - R dl  r = z a d ^ Thus, ECT1026 Field Theory r  z 0 dl 0 -1 0 0 ^ = z dl Step 2 - Segment AC: Along segment AC, dl is perpendicular to R  dl  R = - dl  r = z dl = z a d. ^ ^ ^ ^ ^ dl =a d A C ^ d dl =  a d ^ (opposite direction) ^ r = - R a or r = - R dl  r = z a d ^ ^ Thus,

Circle:  = 2  2r  =    r = (/2)×2r dl  (d/2)×2r = r d circumference Circle:  = 2  2r (Full circle)  =    r = (/2)×2r (Half circle) r d dl  (d/2)×2r = r d r (radius=r) dl =a d (radius=a) Thus,

∫ ò ò moI B = 4 C A ^ B Step 2 - Segment AC: ECT1026 Field Theory Step 2 - Segment AC: Along segment AC, dl is perpendicular to R , and dl  R = z dl = z a d. ^ B = moI 4 l R2 dl  R ∫ ^ A C ˆ m I ò ad f z (R=a) radius = 2 4 p a ^ m I ò ˆ = z df z 4p a B

Example 3.2-3: Magnetic Field of a Circular Loop ECT1026 Field Theory Example 3.2-3: Magnetic Field of a Circular Loop A circular loop of radius a carries a steady current I. Determine the magnetic field B at a point on the axis of the loop. Answer:

z Any element dl on the circular loop is ECT1026 Field Theory Our task is to obtain an expression for B at point P(0,0,z) Any element dl on the circular loop is perpendicular to the distance vector R At the same distance R from point P, with R = √ a2 + z2 z The magnitude of dB due to element dl is given by:

 dB is in the r-z plane, therefore it has components fo dBr and dBz ECT1026 Field Theory Direction of dB is in the plane containing R and dl dB is in the r-z plane, therefore it has components fo dBr and dBz the z-components of B due to dl and dl’ add because they are in the same direction, but Their r-components cancel because their are in opposite directions Consider element dl’ located diametrically opposite to dl  The net magnetic field is along z-axis only

Hence, the net magnetic field is along z-axis only ECT1026 Field Theory Hence, the net magnetic field is along z-axis only For a fixed point P(0,0,z) on the axis of the loop, all quantities are constant, except for dl (Length of circular loop) circumference

Since,  At the center of the loop (z = 0) At point very far away from ECT1026 Field Theory Since,  At the center of the loop (z = 0) At point very far away from the loop such that z2 >> a2