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