# Video: Eg17S1-Physics-Q21B

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?

07:58

### 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, weβve got two wires, π and π. And we know that they are long, that theyβre parallel to each other, and theyβre current-carrying wires. Now, initially, weβve 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 theyβre both long current-carrying wires, they will each be producing their own magnetic field. Specifically, we can recall that if weβve 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, thatβs the direction that the magnetic field lines are pointing in, is given by Ampereβs right-hand rule.

Ampereβs 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 weβll call π΅, is given by multiplying π, thatβs the permeability of the space that the wire is in, by the current through the wire, which weβll call πΌ subscript π€, for the current through the wire. And weβll label that on the diagram as well. And we divide π and πΌ π€ by two π and the distance between the wire and the point at which weβre 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 Ampereβs 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 πβs 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 itβs moved three centimetres closer to π. In other words, this distance is three centimetres. Now before we do anything else, letβs realise that the magnetic field due to π is pointing in this direction. And therefore, at πβs location, it will be pointing into the screen. Thatβs away from us observing this whole situation.

However, πβs magnetic field is pointing in this direction, and so, at πβs 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 πβs magnetic field and πβs magnetic field, we need to remember that one of them should be positive and the other should be negative. It doesnβt matter which is which as long as we account for the fact that theyβre pointing in opposite directions.

So, with that in mind, letβs work out the total magnetic field at point π before the wire π is moved. Letβs call this overall magnetic field π΅ subscript bef, for π΅ subscript before. And this is equal to firstly the contribution due to πβs magnetic field, which will be π multiplied by the current in wire π, which is two πΌ. And we divide this by two π and the distance between π and the point at which weβre considering the magnetic field. So, thatβs this distance here.

But then, that distance is 10 centimetres. Thatβs what weβve 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 πβs magnetic field. But we also need to remember that we said πβs contribution was positive. And so, πβs contribution is going to be negative. So, from πβs contribution we subtract π multiplied by the current in π, thatβs πΌ, and divided by two π and the distance between π and π. Now before we move the wire, thatβs this distance here, which is 10 centimetres, as well.

So, we multiply two π by 10 centimetres in the denominator once again. Now itβs 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 πΌ, thatβs this bit here, minus π multiplied by πΌ, thatβs 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 πΌ becomes two ππΌ. And to π multiplied by 10 centimetres becomes 20π centimetres. Now two ππΌ minus ππΌ is simply equal to ππΌ. So, at this point, weβve worked out the magnetic field at point π before we move the wire π. We can write it down over here. And itβs equal to ππΌ divided by 20π centimetres.

Now letβs work out the magnetic field after we move the wire π, which weβll call π΅ subscript aft. Well, this is going to be equal to firstly the contribution due to πβs magnetic field, which is not changing. So, once again, we get π multiplied by two πΌ. Thatβs the current in wire π divided by two π times 10 centimetres. And then, from this we need to subtract πβs contribution.

But, this time, π is a lot closer to π. In fact, itβs three centimetres closer. So, now the distance weβre considering is going to be between π and the wire itself, which is 10 minus three, or seven centimetres. And itβs worth remembering that in this time the current through the wire is not said to be changing, itβs only the distance between π And the wire. So, the contribution from πβs field is π multiplied by πΌ, thatβs the current in π. And we divide this by two π and seven centimetres now.

Now, at this point, weβll do things slightly differently. Weβll say that the twos in the numerator and denominator of the first fraction cancel. So, the first fraction becomes ππΌ 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 ππΌ 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 ππΌ 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 ππΌ 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, weβre left with seven ππΌ minus five ππΌ divided by 70π centimetres. This becomes two ππΌ 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, weβve come to the conclusion that the magnetic field at point π after we move the wire π is given by ππΌ 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 ππΌ divided by 20π centimetres. However, after we move the wire, we still have ππΌ 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, weβve 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 weβve 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.