25

Magnetic field

Field due to current in a long straight wire

AS Shown in Chapter 23 (p. 198) the field strength,• called the magnetic density, is given by

where I is the current through the straight wire and B is the resulting flux density at a Bl)int distance d from the wire. equals Ixrmcabiiity Of the medium. The lines Of force Of this field are circles centred upon the Wire (as stated in Chapter 23), and this is shown in Fig. 23.2. The directions of the lines of force are given by the 'corkscrew rule' iKCOrding to which these directions are c:lœkwise when one looks along the Wire in the direction Of the current.

Fig.

calculations

Neutral points in magnetic fields

If a magnetic field results from more than one current-carrying conductor or magnet then at a certain place in the field the densities may be equal in magnitude and opposite in direction so that their effects cancel, the resultant flux density is Such a place is called a 'neutral IXú1t'.

Example 1

A long. straight, vertical wire carries a downward current Of 4.0 A. The earth's magnetic field in Which this wire is placed has a horizontal component or 1-6 T. Calculate:

(a) the resultant horizontal magnetic flux density at a

I Ocm to the west Of the wire

(b) the distance from the Wire of the neutral point.

Method

(a) The flux density due to the wire at a d of IOcm (0.10m) is given by x 10

Due west Of the Wire this flux density is, accxM-ding to the corkscrew rule, directed northwards. It therefore adds to the earth's horizontal flux density. So the resultant flux density is If, x x UV^S •r or 24/'T.

(b) At the neutral the flux density due to the Wire is equal in (o the Of the earth's field.

(a) 24"T, (b) 5.0cm

209

CALCULATIONS A-LEVEL PHYSICS A vertical wire carries a downward current Of 12cm east Of this there is another Vcrtical wire carrying an equal downward current. The earth's horizontal component is 1.6 10 •ST.

Within a toroid (an endless coil. scc FiB 25.2) the magnetic flux density is given by

A solenoid is a long coil, i.e. its length is considerably greater than its diameter, as shown

F'. 25.3 A

A solenoid can thought of as part of a large toroid. and the turns of the remainder of the toroid are far from the middle Of the solenoid to affect the flux density there. Hence the same formula (25.2) applies to a solenoid.

Example 2

Calculate the flux density in the               of a

having 10 turns                          and carrying a Current

Of 0.50 A. The medium within the sokncid is air. for which the is x 10 -7 H m - ^I Method

The flux density is given Equatk:m 25-2:

210

What is the flux detBity at a distance 2.0cm from the first Wire and IOOcm from     Other?

2  Two long parallel. straight wires are IOctn apart. One wire carries a current of 2.0A and the Other carries 3.0A. In the resulting magnetic field there is a neutral point. Calculate its distance from the

2.0 A wire

(a) When the currents are in the same direction

(b) when the currents are in opposite directions.

A having turns per mctrc and carrying a Current Of 0.050A lies with its axis east•CSt. Well inside the solenoid is a small needle points 37' west of north. Calculate the Earth's horizontal magnetic field

An air-cored toroid has 2m turns and a length of 15cm. Around its centre is wound a coil of

3.0cm with 20 turns. If the current in the toroid is initially 20 and is reduced Steadily to Zero in a time Of O, IOS. what EMF Will be induced in the 20•turn coil during this time. (Take permeability of air to x 10-7 Hm -l .) Exercise 25.2:

Examination questions

I flux density at a certain rx'int P to a long. straight wire carrying a current I is 3.0mT. A lirx perpendicular to the wire and EVEssing through P mccts a rx_'int Q which is twice as far from the wire as P. What the flux density at O When current in the Wire is reduced to OS I?

2 A slinky spring Of turns is stretched uniformly along a horizontal bench-top WI-xn a current of 1_20 A is the spring. it acts as a

(a) Calculate the magnetk nux &nsity at centre of this sokruSd when the tension in the spring is such that its length is m.

(b) The tension in the spring is that its length find the new flux density at centre Of tir

ICCE-A

       Use                  equations F BIL B

           E —L to              that henry        metre (H m -l )

MAGNETIC FIELD CALCULATIONS

(i) Calculate the flus density BC of magnetic field at P due to the current in the cable. Take the Of air to be 1.3 x

(ii)On Fig. 25 A draw an arrow at P to show the direction of BC.

Of the x to -S T. Assume

Of the magnetic field that this acts in the direction Of truc North. and that there are no other magnetic fields apart frŒn that of the Current in cable.

Calculate the resultant horizontal magnetic flux density B at P. and state the direction in whkh the compass points, when the carriage

is an appropriate unit for g.          henry is the unit                      is oriented         its front:

for self•indtxtanee L.) L in BIL denotes length. (i) touards the cast;   4 (a) Fig 25.4 shows a rear-view cross-section of the Of a railway carriage and Of an ( ii) towards the north. (OCR