Lecture 16

Magnetism (3)

HistoryHistory

1819 Hans Christian Oersted discovered that a compass 1819 Hans Christian Oersted discovered that a compass needle was deflected by a current carrying wireneedle was deflected by a current carrying wireThen in 1920s Jean-Baptiste Biot and Felix Savart Then in 1920s Jean-Baptiste Biot and Felix Savart performed experiments to determine the force exerted performed experiments to determine the force exerted on a compass by a current carrying wireon a compass by a current carrying wire

Magnetic Field of a long straight wireMagnetic Field of a long straight wire

I=0 I

Biot & Savart’s ResultsBiot & Savart’s Results

dBdB the magnetic field produced by a the magnetic field produced by a small section of wiresmall section of wire

dsds a vector the length of the small a vector the length of the small section of wire in the direction of the section of wire in the direction of the currentcurrent

rr the positional vector from the section the positional vector from the section of wire to where the magnetic field is of wire to where the magnetic field is measuredmeasured

I the current in the wireI the current in the wire

angle between angle between dsds & & rr

r

ds

dB

Biot & Savart’s ResultsBiot & Savart’s Results

dBdB perpendicular to perpendicular to dsds

||dBdB| inversely proportional to || inversely proportional to |rr||22

||dBdB| proportional to current I| proportional to current I

||dBdB| proportional to || proportional to |dsds| |

||dBdB| proportional to sin | proportional to sin

Biot–Savart LawBiot–Savart Law

All these results could be All these results could be summarised by one “Law”summarised by one “Law”

Putting in the constant

Where 0 is the permeablity of free space

Magnetic Field from Biot-Savart LawMagnetic Field from Biot-Savart Law

BB = dB = dB11+dB+dB22+…+dB+…+dBii

I.e. I.e. BB = =dBdB

r1

ds1

dB1

r2

ds2

dsi

dBi

ri

dB2

We can use the Biot-Savart law to calculate the magnetic field due to any current carrying wire

Magnetic Field due to CurrentsMagnetic Field due to Currents

The passage of a steady current in a wire produces a The passage of a steady current in a wire produces a magnetic field around the wire.magnetic field around the wire.

Field form concentric lines around the wireField form concentric lines around the wire Direction of the field given by the right hand rule.Direction of the field given by the right hand rule.

If the wire is grasped in the right hand with the thumb in the If the wire is grasped in the right hand with the thumb in the direction of the current, the fingers will curl in the direction of direction of the current, the fingers will curl in the direction of the field.the field.

Magnitude of the field Magnitude of the field I

Magnitude of the field I

r

B

o called the permeability of free space

[Q]:[Q]:

The two wires in the figure below carry currents of 3.00A and 5.00A in The two wires in the figure below carry currents of 3.00A and 5.00A in the direction indicated. Find the direction and magnitude of the the direction indicated. Find the direction and magnitude of the magnetic field at a point midway between the wires. magnetic field at a point midway between the wires.

3.00 A 5.00 A

20.0 cm

Magnetic Field of a current loopMagnetic Field of a current loop

Magnetic field produced by a wire can be enhanced Magnetic field produced by a wire can be enhanced by having the wire in a loop.by having the wire in a loop.

x1

I

x2

B

N loops Current NI 1 loopCurrent I

[Q] What is the magnetic field at point Q in Fig.?

[Q] What is the magnitude and direction of the magnetic field at point P in Fig.?

[Q] Use the Biot-Savart Law to calculate the magneticfield B at C, the common center of the semicircular areaAD and HJ of radii R1=8 cm and R2=4 cm, forming part ofthe circuit ADJHA carrying current I=10 A, as seen figure.

Ampere’s Law

Consider a circular path surrounding a current, divided in segments l, Ampere showed that the sum of the products of the field by the length of the segment is equal to o times the current.

I

r

Bl

Consider a case where B is constant and uniform:

Then one finds:

Magnetic Force between two parallel conductorsMagnetic Force between two parallel conductors

l

d

1

2 F1

B2 I1

I2

Force per unit length

Definition of the SI unit AmpereDefinition of the SI unit Ampere

If two long, parallel wires 1 m apart carry the same current, and the If two long, parallel wires 1 m apart carry the same current, and the magnetic force per unit length on each wire is 2x10magnetic force per unit length on each wire is 2x10-7-7 N/m, then the N/m, then the current is defined to be 1 A.current is defined to be 1 A.

Used to define the SI unit of current called Ampere.

ExampleExample

Two wires, each having a weight per units length of 1.0x10Two wires, each having a weight per units length of 1.0x10 -4-4 N/m, are N/m, are strung parallel to one another above the surface of the Earth, one strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 mm what must be the current in each in distance of separation is 0.10 mm what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current).wires carry the same current).

l

d

1

2I1

I2

l

d

1

2

F1

B2I1

I2

mg/l

Magnetic Field of a solenoidMagnetic Field of a solenoid

Solenoid magnet consists of a wire coil with multiple Solenoid magnet consists of a wire coil with multiple loops.loops.

It is often called an electromagnet.It is often called an electromagnet.

Solenoid MagnetSolenoid Magnet

Field lines inside a solenoid magnet are parallel, uniformly spaced Field lines inside a solenoid magnet are parallel, uniformly spaced and close together.and close together.The field inside is uniform and strong.The field inside is uniform and strong.The field outside is non uniform and much weaker.The field outside is non uniform and much weaker.One end of the solenoid acts as a north pole, the other as a south One end of the solenoid acts as a north pole, the other as a south pole.pole.For a long and tightly looped solenoid, the field inside has a value:For a long and tightly looped solenoid, the field inside has a value:

Solenoid MagnetSolenoid Magnet

n = N/n = N/ll : number of (loop) turns per unit length. : number of (loop) turns per unit length.

I : current in the solenoid.I : current in the solenoid.

Example:Example:

Consider a solenoid consisting of 100 turns of wire and Consider a solenoid consisting of 100 turns of wire and length of 10.0 cm. Find the magnetic field inside when it length of 10.0 cm. Find the magnetic field inside when it carries a current of 0.500 A.carries a current of 0.500 A.