Magnetism

Lecture 3.12

Electromagnetic Induction

Applications

Magnetic force

Magnetic force A wire carrying a current in an external magnetic field.

Force on the wire is perpendicular to the plane containing the line of the wire and the line of the magnetic field. i.e. into the page

N N S S

I

F

F

Current and magnetic field are perpendicular to each other as shown.

B = F/IL (B perpendicular to I)

Force (F) on the wire is proportional to the current and the length (L) of the wire in the magnetic field

B is the magnetic field strength

units of B are Newtons per metre per ampere which is called Tesla (SI units)

1T = 1N.m-1.A-1

N N S S

I

F

F

F IL∝

Magnetic force

sinF qvB θ=

( )( )LF It B BILt

= =

090θ =F qvB=

1

NCms−

A wire carrying a current in an external magnetic field. Current and magnetic field are not perpendicular but at angle θ to each other.

Force on the wire is proportional to the component of the magnetic field that is perpendicular to the wire

N N S S I F

B

θ

θ θ

Sinθ = X/B

F = (Bsinθ).IL

If θ =90o then as before F= BIL

If θ = 0 (B and I are parallel) then F = 0

Magnetic force

Magnetic Force

Application: Audio speaker

A wire carrying a current in an external magnetic field experiences a force

Magnetic force on the current-carrying coil. converts electrical signal to sound wave

Example Speaker coil with 50 turns of wire and diameter 2 cm. positioned in magnetic field of 0.1 T. If the current in the coil is 1.5 A and we assume that the magnetic field is perpendicular to the entire coil, calculate (a) the force acting on the coil and cone and (b) the corresponding acceleration if the mass of the coil and cone is 0.015 Kg.

F = (Bsinθ).IL θ =90o therefore F= BIL

L = N x circumference = N (πd)

50 0.02 3.14L m mπ= × × =

0.1 1.5 3.14 0.47F T A m N= × × =

F BIL=

20.47 31.40.015

F Na msm kg

−= = =

Advantages •Low maintenance •no contact with the ground •No moving parts •trains float, •no friction •extremely fast 500km/h

•Electro-magnets run the length of the guideway •No conventional engine or wheels •Magnetically propelled along the guideway •floats on a magnetic field

Disdvantages •costly •existing infrastructure

Magnetic Levitation trains

Magnetic force

This effect is used in generators to produce ac current (power plants, etc)

Magnetic Induction Moving charges (current) cause a magnetic field

A changing magnetic field in the vicinity of a wire or coil will induce a current in the wire or coil

N S

I

Magnetic Flux

Measure of the number of magnetic field lines that cross a given area A.

Φ =BA

Φ =(BCos θ)A

Magnetic flux Φ

θ BCos θ

B θ

θ = 0 Cosθ = 1 Φ =BA

θ = 900 Cosθ = 0

Φ =0

A = area of coil

B

B

B

Electromagnetic Induction

A changing magnetic field in the vicinity of a wire or coil will induce a current in the wire or coil

Faraday’s law of magnetic induction

Induced voltage is proportional to the rate of change of magnetic flux through the coil.

emf Nt

∆Φ= −

∆

- + ammeter

0.0 98

N =number of turns in coil

Secondary circuit

Primary circuit

emf Nt

∆Φ= −

∆

A bar magnet is completely removed in a period of 0.5s from its position close to a coil of radius 5cm with 50 turns as shown. Initially the magnetic field B over the area of the coil is of strength 50 mT. Determine the induced emf in the coil.

( )cosB AθΦ =

N S B

BAΦ =0θ =

( ) ( )23 2 7 250 10 5 10 3927 10 .T T mπ− − −Φ = × × = ×

( )77

3927 10 07854 10

0.5V

t s

−−

× −∆Φ= = ×

∆

7

2

50 7854 10

3.927 10

emf N Vt

emf V

−

−

∆Φ= − = × ×

∆= − ×

Magnetic Induction

diaphragm Stationary magnet

sound

Microphone coil moves with diaphragm

induced current corresponds to sound

magnet

Pickup coil N S S N

amplifier

Guitar string

Electric guitar

Current induced in pickup coil by vibrating magnetised string

seismograph

Applications Magnetic flux through the coil changes

Oscillating string changes magnetic flux in the coil

Electromagnetic Induction

Transformer

Iron

ac Voltage in ac Voltage out

Charger: mobile phone I pod

ac voltage

X-ray machine—high voltage

Application

P=IeffVeff

Transformers important in power transmission

Voltage out is proportional to the number of turns in the secondary coil p p

s s

V NV N

=

Secondary circuit

Primary circuit

1/2mv² = qV 2qVvm

=

P=Ieff2R

p p

s s

V NV N

=

example

A device (“bug zapper”) requires 4400V to operate. If the household mains voltage is 220V, how many turns must the secondary coil of the transformer have if the primary coil has 25 turns.

ss p

p

VN NV

=

440025 500220s

VNV

= =

Electromagnetic Induction

Airport metal detectors

Passengers subjected pulsed magnetic field

Current induced in metal object due to pulsed magnetic field

Passengers walk through arch

Passenger with metal object

Metal object produces its own magnetic field

Magnetic field of object detected

Two coil metal detector Essentially works on same principle

Application

Induction loop

Magnetic fields, other Applications

stores digitally encoded information in magnetic format

Computer hard-disk drive

Disks spin, magnetic read/write heads move across the disk and can detect and modify the magnetization of the material immediately under it

Read/write head

Magnetoencephalography (MEG) Study of magnetic fields produced by electrical activity of the brain

Provides information on spontaneous brain function or that evoked by auditory or visual stimuli

Electromagnetic Induction

Even thoughts result in magnetic fields that can be detected outside head.

Magnetic fields 10-13 T

MEG can pinpoint the source of the brain activity

MEG and MRI

ac Electricity Generator

N S N S

emf induced in the coil by changing the magnetic field through the coil

Converts mechanical to electrical energy

Mechanical energy supplied by: •Falling water: •Expanding steam:

Hydroelectric dam Coal/oil fired station

Alternating current -V0

V

0 t 0

Induced emf changes sign

Electric Motor

Converts electrical energy into mechanical energy

ac current in

N

S

Magnetic fields can penetrate tissue with little or no adverse effects ---can be used to probe the body

Medical uses of Magnetic fields

Main medical application

Nuclear Magnetic Resonance NMR Magnetic Resonance Imaging MRI

Like electrons, protons (positively charged) in the nucleus also act as though they are spinning about an axis through their centres. Spinning protons act like a current loop and so creates a tiny magnetic field

Nuclei of some atoms (particularly Hydrogen) have small magnetic fields

In strong magnetic field they align with the field.

Medical uses of Magnetic fields MRI

Non-invasive imaging technique that discriminates between body tissues.

Patient in strong magnetic field. Radio frequency signal applied (reorients the proton spin). Absorption and re-emission measured and computer image generated showing the types of tissue present.

diagnostic tool for soft tissue – organs, ligaments, the circulatory system, spinal column, brain

MRI uses superconducting magnet;

– 6 x104 times the Earth’s magnetic field. Earth’s magnetic field ≈ 0.5 x10-4tesla. fridge magnet ≈ 10-3 tesla.

MRI scanner magnet ≈ 3 tesla

A horizontal wire lying in the east-west direction has a mass per length of 0.18g/metre and carries a current I. The wire is in a northward-directed horizontal magnetic field of strength 0.5 Tesla. Find the minimum current (I) such that the magnetic force on the wire can support the weight of the wire.

I = 3.53mA

Magnetic force F = B IL= mg

I

Gravitational force =mg

Magnetic force Magnetic field

F = 0.5T. I. L = 0.18x10-3kg. L . 9.8 ms-2

I = 0.18x10-3kg. 9.8 ms-2

0.5T. = 3.53 x 10-3A

Example

A rectangular loop is placed in a uniform magnetic field with the plane of the loop perpendicular to the direction of the field. If a current is made to flow through the loop in the sense shown by the arrows, the magnetic field exerts on the loop: 1. a net force. 2. a net torque. 3. a net force and a net torque. 4. neither a net force nor a net torque.

A wire placed in magnetic field such that its length is perpendicular to the field that is of strength 10 milliTesla. If the current in the wire is 15 milliamps and the magnitude of the force on the wire is 30x10-6 N, what is the length (in cm) of the wire in the magnetic field?

B = F/IL

Therefore L =F/BI = 30x10-6 N

(10x10-3 T) (15x10-3 A)

L = 2x10-1m = 0.2m = 20cm.

Example

A wire placed in a uniform magnetic field such that it lies at an angle of 30o to the field which is of strength 10milliTesla. If the current in the wire is 15 milliamps and the length of the wire in the magnetic field is 5cm, calculate the force on the wire.

Example

F = Bsinθ IL

F =(10x10-3 T) Sin30o (15x 10-3A)(5x10-2m)

F= 3.75x10-6Newtons

F = Bsin900 IL

F =(10x10-3 T) Sin90o (15x 10-3A)(5x10-2m)

F =(10x10-3 T) (15x 10-3A)(5x10-2m)

F =7.5x 10-6N

Perpendicular magnetic field