### Video Transcript

In the figure, π and π are long,
parallel, current-carrying wires. If wire π is displaced three
centimetres towards point π, what will happen to the total magnetic flux density at
π?

Okay, so, in this question, weβve
got two wires, π and π. And we know that they are long,
that theyβre parallel to each other, and theyβre current-carrying wires. Now, initially, weβve been told
that both π and π are 10 centimetres away from point π. π is 10 centimetres to the left
and π is 10 centimetres to the right.

Now because theyβre both long
current-carrying wires, they will each be producing their own magnetic field. Specifically, we can recall that if
weβve got a long straight current-carrying wire, then a circular magnetic field will
be produced at every point along the length of the wire. And the direction of the magnetic
field, thatβs the direction that the magnetic field lines are pointing in, is given
by Ampereβs right-hand rule.

Ampereβs right-hand rule tells us
that if the thumb on our right hand is pointing in the same direction as the current
in the current-carrying wire, then the direction in which the fingers on our
right-hand curl will give us the direction of the magnetic field. And we could even go a step further
by working out the magnetic field at a particular point that we can say is a
distance π away from the wire.

Now in a situation like this, the
strength of the magnetic field produced by a long straight current-carrying wire,
which weβll call π΅, is given by multiplying π, thatβs the permeability of the
space that the wire is in, by the current through the wire, which weβll call πΌ
subscript π€, for the current through the wire. And weβll label that on the diagram
as well. And we divide π and πΌ π€ by two
π and the distance between the wire and the point at which weβre calculating the
magnetic field.

So, using this information, we can
realise that each of the wires in our question, π and π, produce their own
magnetic field with the directions given by Ampereβs right-hand rule. And moreover, we can consider each
of the magnetic field lines produced by π and π that intersect the point π. And so, we can work out the total
magnetic field at π by summing up the individual contributions from π and πβs
magnetic fields.

So, we can work out the magnetic
field at π before π is moved. And we can work out the magnetic
field at π after itβs moved three centimetres closer to π. In other words, this distance is
three centimetres. Now before we do anything else,
letβs realise that the magnetic field due to π is pointing in this direction. And therefore, at πβs location, it
will be pointing into the screen. Thatβs away from us observing this
whole situation.

However, πβs magnetic field is
pointing in this direction, and so, at πβs location, will be pointing towards us,
the observers. And so, the effects of these two
magnetic fields will work against each other. In other words, when we sum the
contributions due to πβs magnetic field and πβs magnetic field, we need to
remember that one of them should be positive and the other should be negative. It doesnβt matter which is which as
long as we account for the fact that theyβre pointing in opposite directions.

So, with that in mind, letβs work
out the total magnetic field at point π before the wire π is moved. Letβs call this overall magnetic
field π΅ subscript bef, for π΅ subscript before. And this is equal to firstly the
contribution due to πβs magnetic field, which will be π multiplied by the current
in wire π, which is two πΌ. And we divide this by two π and
the distance between π and the point at which weβre considering the magnetic
field. So, thatβs this distance here.

But then, that distance is 10
centimetres. Thatβs what weβve been told in the
diagram. So, we multiply two π in the
denominator by 10 centimetres. And then, to this, we need to add
the contribution due to πβs magnetic field. But we also need to remember that
we said πβs contribution was positive. And so, πβs contribution is going
to be negative. So, from πβs contribution we
subtract π multiplied by the current in π, thatβs πΌ, and divided by two π and
the distance between π and π. Now before we move the wire, thatβs
this distance here, which is 10 centimetres, as well.

So, we multiply two π by 10
centimetres in the denominator once again. Now itβs at this point that we can
see that both denominators are the same. And therefore, we can add the
numerators, or in this case subtract one from the other. As a result, we find that π΅ sub
bef is equal to π multiplied by two πΌ, thatβs this bit here, minus π multiplied
by πΌ, thatβs this bit here, all divided by the common denominator, which is two π
multiplied by 10 centimetres.

At which point, we can expand and
simplify. π multiplied by two πΌ becomes two
ππΌ. And to π multiplied by 10
centimetres becomes 20π centimetres. Now two ππΌ minus ππΌ is simply
equal to ππΌ. So, at this point, weβve worked out
the magnetic field at point π before we move the wire π. We can write it down over here. And itβs equal to ππΌ divided by
20π centimetres.

Now letβs work out the magnetic
field after we move the wire π, which weβll call π΅ subscript aft. Well, this is going to be equal to
firstly the contribution due to πβs magnetic field, which is not changing. So, once again, we get π
multiplied by two πΌ. Thatβs the current in wire π
divided by two π times 10 centimetres. And then, from this we need to
subtract πβs contribution.

But, this time, π is a lot closer
to π. In fact, itβs three centimetres
closer. So, now the distance weβre
considering is going to be between π and the wire itself, which is 10 minus three,
or seven centimetres. And itβs worth remembering that in
this time the current through the wire is not said to be changing, itβs only the
distance between π And the wire. So, the contribution from πβs
field is π multiplied by πΌ, thatβs the current in π. And we divide this by two π and
seven centimetres now.

Now, at this point, weβll do things
slightly differently. Weβll say that the twos in the
numerator and denominator of the first fraction cancel. So, the first fraction becomes ππΌ
divided by 10π centimetres. And then, we can expand the
denominator of the second fraction. So, we have two π multiplied by
seven centimetres, which will result in us having ππΌ divided by 14π
centimetres. So, now we do not have a common
denominator, which means we need to find one.

And we can do this if we multiply
the first fraction by seven divided by seven and the second fraction by five divided
by five. Now the reason we do this is
because multiplying by seven divided by seven is the same thing as multiplying by
one. But this time the first fraction
becomes seven ππΌ divided by 70π centimetres. And the second fraction, when we
multiply by five divided by five, same thing as multiplying by one once again,
becomes five ππΌ divided by 70π centimetres again.

So, now we found a common
denominator, 70π centimetres, which means that now we have a common
denominator. And we can add, or in this case
subtract one numerator from the other. So, weβre left with seven ππΌ
minus five ππΌ divided by 70π centimetres. This becomes two ππΌ divided by
70π centimetres. But then, we can remember that 70
is equal to two times 35. And so, this factor of two will
cancel with this factor of two.

So, at this point, weβve come to
the conclusion that the magnetic field at point π after we move the wire π is
given by ππΌ divide by 35π centimetres. So, now we have the magnetic field
before we move the wire π and after we move the wire π. This means that we can compare
these two quantities.

Now before we move the wire, now
before we move the wire, we had the field ππΌ divided by 20π centimetres. However, after we move the wire, we
still have ππΌ in the numerator. But this time, we have a larger
value, 35π centimetres in the denominator. So, if we have the same quantity
divided by a larger value in the denominator, then this means that the value of π΅
aft is going to be smaller than π΅ bef. Or, in other words, π΅ aft, the
magnetic field at π after we move the wire π, is less than the π΅ bef, the
magnetic field before we move the wire π.

And so, at this point, weβve
reached the answer to our question. If wire π is displaced three
centimetres towards point π, we can say that the total magnetic flux density, or
the magnetic field strength, which is what weβve been calculating, ends up
decreasing. Because as we saw, the magnetic
field at point π after we move the wire is less than the magnetic field at point π
before we move the wire.