Mod-7 Biotamp Law

Biot and Savart law The magnitude of magnetic field B at due to a current-length element i dl a distance of r from fro i the element is given by dB points µo i dl sin θ into the page dB = 2 4π r where θ is the angle between the length element dl and the position vector r which extends from the current element to the point at which dB is r measured. Here µo is a constant called permeability constant dl whose value is defined to be exactly µo = 4 π x 10-7 T.m /A in vector form  µo i dl × r dB = 4π r 3 The vector equation and its scalar form are known as the law of Biot and Savart. The direction of B due to a current element is important feature of the above equation. If both the current and the point at which B is measured are situated on the plane of the page then the direction of the vector product i.e. the direction of dB should point normally in to the plane of the page.

Magnetic field due to a straight current carrying wire At a distance of R on either side of the wire carrying current i is given by

Right side

Left side

R +∞

µ

o B = ∫ dB = 4π −∞

+∞

µo i i dl sin θ = 2 ∫ r 2π R −∞

B points normally out of the plane

R B points normally into the plane

The lines of B forms a closes circular loop around the straight wire as shown in figure below

Ampere’s Law The mathematical statement of Amperes law is given by   B ∫ • dl = µo i The circle on the integral sign means that the scalar product B.dl is to be integrated over a closed path called Amperian Loop. The current I on the right hand side is the net current encircled by the loop. In words the law can be stated as " The integral of B.dl around any closed mathematical path equals u0 times the current intercepted by the area spanning the path " Explanation : Figure below shows three long straight wires currents i1 , i2 and i3 either directly into or directly out of the page. An arbitrary Amperian loop encircles the two currents   but not the third. The integral ∫ B • dl = B dl cosθ can be evaluated just by determining the net current encircled by the loop. To do this we need to follow the following sign convention.

i3

i1

i2

“ Curl the right hand fingers in the direction of integration along the Amperian loop. Current passing through the loop in the general direction of outstretched thumb is assigned positive sign and vice-versa.” With this convention current i1 is positive and i2 is negative so that net current encircled by the loop is ienc = i1 - i2

Magnetic field near a straight Wire carrying a current. Let us consider a circular amperian loop of radius R centering the axis of a wire carrying current i. The figure shows the cross-section of a wire running normally out of the page. The general direction of B at any point of the circular in tangential to the path and is anticlockwise. So the dot product B.dl at any point of the path is equal to B dl. As B has constant value over the circular path the closed path integral is   B B ∫ • dl = ∫ B dl = B ∫ dl = B(2πR) Now as the current enclosed by the path is I we can write from Ampere’s law B (2πR) = µoi µo i or , 2π R

R R

Magnetic field inside a current carrying wire Figure below shows the cross-section of a wire of radius R carrying current i. To determine the magnetic field B inside the wire let us consider a circular loop of radius r (r,R). The general direction of B at any point of the circular path is tangential to the path and is anticlockwise. So the dot product B.dl at any point of the path is equal to B dl. As B has constant value over the circular path the closed path integral is  B ∫ • dl = ∫ B dl = B ∫ dl = B(2πr ) Now as the current enclosed by the path is   i = ∫ J • dA where J is the current density. Considering J to be constant over the cross-section of the wire the surface integral i is equal to J(π r2). Now as J = π R2 i i r2 2 i enc = 2 (π r ) = πR R2

πr2

r πR2

R

Now from Amperes Law we get i r2 B (2πr) =µo 2 R µo i r or, B = π R2 So Magnitude of B is directional to the distance from the axis of the wire at positions inside the wire.

Magnetic field inside a Solenoid A solenoid is a long tightly wound helical coil of wire whose diameter is small compared to its length. The magnetic field generated in the centre, or core, of a current carrying solenoid is essentially uniform, and is directed along the axis of the solenoid. Outside the solenoid, the magnetic field is far weaker. The figure below shows (rather schematically) the magnetic field generated by a typical solenoid. The solenoid is wound from a single helical wire which carries a current i . The winding is sufficiently tight so that each turn of the solenoid is well approximated as a circular wire loop, lying in the plane perpendicular to the axis of the solenoid. Suppose that there are n such turns per unit axial length of the solenoid. A stretched out solenoid and the schematic diagram of it is shown in figure (a) and (b) respectively.

In order to find the magnetic field inside the solenoid let up apply Ampère's circuital law to the rectangular loop abcd. We must first find the line integral of the magnetic field around abcd. Along bc and da the magnetic field is essentially perpendicular to the loop, so there is no contribution to the line integral of the field from these sections of the loop. Along cd the magnetic field is approximately uniform, of magnitude B, say, and is directed parallel to the loop. Thus, the contribution to the line integral from this section of the loop is BL, where L is the length of cd. Along ab the magnetic field strength is essentially negligible, so this section of the loop makes no contribution to the line integral. It follows that the line integral of the magnetic field around abcd is simply W = BL By Ampère's law, this line integral is equal to µo times the algebraic sum of the currents which flow through the loop abcd. Since the length of the loop along the axis of the solenoid is L, the loop intersects nL turns of the solenoid, each carrying a current i. Thus, the total current which flows through the loop is Σi = nLi . Ampère's law yields BL = µo nLi which reduces to B = µo ni This, result is exact in the limit in which the length of the solenoid is very much greater than its diameter

Problem sheet 1. A loop of a wire has the shape of two concentric semicircles connected by two radial segments. The loop carries current I as shown in the figure. Find the magnetic field at point P using law of Biot –Savart

P 2.

Use Biot-Savart law to calculate the magnetic field at the center of the two concentric arcs forming a closed loop. The current is I and the angle made at the center is 900. a I

b C

3. 4. 5

6. 7. 8.

The magnitude of the magnetic field 88.0 cm from the axis of a long straight wire is 7.30 µT. what is the current through the wire? A 10 gauge bare copper wire 92.6 mm in diameter)can carry a current of 50 A without overheating. For this current what is the magnetic field at the surface of the wire A surveyor is using a magnetic compass 20 ft below a power line in which there is a steady current of 100 A. What is the magnetic field at the site of the compass due to the power line? Two parallel wires are 10.0 cm apart. What equal currents must be in the wires if the magnetic field halfway between them is to have a magnitude of 3.0 mT? Answer for antiparallel currents. Two parallel wires a distance of d apart carry currents i and 3i in the same direction. Locate point or points where magnetic field is zero. Two long straight wires a distance d apart carry equal antiparrel currents i as in figure below. (a) Show that the magnitude of the magnetic field at point P, equidistant from the wires s given by 2 µoid d R B= π 4R 2 + d 2 (b) In which direction B points? each of the eight conductors in figure below carry 2.0A current into or out of the page. Two paths are   indicated for the line integral ∫ B • dl . What is the path integral for the path (a) at the left and (b) at the right?

(

9

10. 11.

12 13 14

)

Figure below shows the cross section of long cylindrical wire of radius a, carrying a uniformly distributed current i. Assume that 2 = 2.0 cm, i = 100 A and plot B(r) in the range 0
P

15

A solenoid has a length of 1.2 m and a mean diameter of 3.0 cm. It has 5 layers of windings each having 850 turns. If the solenoid current is 5 A calculate (a) the magnetic field near its center. (b) Magnetic flux through the cross-section of the solenoid near its center.

16

Each of two straight parallel wires 10 cm apart carries a current of 100A. Figure below shows a cross-section with the wires running perpendicular to the plane of the page and point P lies on the perpendicular bisector of the line between the wires. Find the magnitu8de and direction of the magnetic field at p when the current in the left hand wire is out of the page and the current on the right hand wire is (a) out of the page and (b) into the page P

17

Figure below shows three parallel wires in the xy plane. Each carries a current of 4.0 A in the positive x direction the separation between the adjacent wires is d = 5.0 cm. What is the magnetic force per meter of wire number (2) in the figure. z

1 x

2

3

y