11.3 Calculating the size of the force produced by the motor effect using the equation F = BIL

The mathematics of the 'kicking wire'!

The size of the force on the conductor depends on:

The magnetic flux density (B) - the closer together the lines of force, the greater the field strength, the greater the resulting force.

The current in the conductor (I) - the greater the rate of charge flowing, the stronger the magnetic effect, the greater the resulting force.

the length of conductor (L) (e.g. copper wire) in the magnetic field - if the force operates over a greater length, so overall a greater force is exerted on the wire.

The net force (F) is proportional to all these three factors combine into one simple equation:   F = B I L

The diagram illustrates the variables in the equation to calculate the force acting on a current carrying wire.

Note that I've made the left of the diagram ~match the application of Fleming's left-hand rule.

To calculate the size of the force

For a conductor at right angles to a magnetic field and carrying a current:

force = magnetic flux density × current × length,    F = B I L

force F, in newtons, N;       magnetic flux density, B, in tesla, T

current, I, in amperes, A;    length, L, in metres, m  (watch out to convert from cm to m = cm/100))

Rearrangements: B = F / IL,  I = F / BL  and  L = F / BI,  and in ANY calculation, watch out to match the units!

You also need to be able to use Fleming's left-hand rule to predict the direction of the resultant force.

 

Examples of calculations

Q1 A 10 cm length of wire carrying a current of 8.0 A is at right angles to a magnetic field of strength 0.25 Tesla.

(a) If the directions of the magnetic field and current are in the plane of the screen, deduce the direction of the force on the wire.

From Fleming's left-hand rule you should deduce the direction of force-motion is directly towards you!

(b) Calculate the size of the force on the wire.

F = BIL = 0.25 x 8.0 x (10/100) = 0.2 N

 

Q2 A 5.0 cm length of wire carrying a current of 3.0 A experiences a force of 5.0 N.

Calculate the magnetic flux density around the wire.

F = BIL,  B = F/IL,   B = 5.0/(3.0 x 5.0/100) = 3.3 T

 

Q3 What current must flow through a 1.5 m wire for it to experience at 90^o a force of 9.0 N in a magnetic flux density of 3.0 T?

F = BIL,   I = F/BL,   I = 9.0/(3.0 x 1.5) = 9.0/4.5 = 2.0 A

 

Q4 In cm, what length of wire carrying a current of 20 A, will experience a force of 25 N at 90^o to a magnetic flux density of 5.0 T?

F = BIL,  L = F/BI,  L = 25/(5 x 20) = 25/100 = 0.25 m = 25 cm

 

Q5 Calculate the force on a 50 m stretch of telephone wire carrying a current of 50 milliamps and the Earth's magnetic field flux acting on the wire is 40 000 nanotesla.

B = 40 000 x 10^-9 = 4.0 x 10^-5 T,  I = 50 x 10^-3 = 5.0 x 10^-2 A,  L = 50 m

F = B x I x L = 4.0 x 10^-5 x 5.0 x 10^-2 X 50 = 1.0 x 10^-4 N

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INDEX physics notes: motor effect of an electric current and applications