Video Transcript
A loop of wire has a potential
difference of 1.2 volts across it. The loop has a self-inductance of
125 millihenries. How much time is required for the
loop to increase the current through it by 0.25 amperes. Give your answer to two decimal
places.
This question is asking us about an
increasing current in a loop of wire. We’re given values for the
potential difference across the loop and the self-inductance of the loop. We’re asked to work out how much
time it takes for the current to increase by an amount of 0.25 amperes. Now, there’s a formula that’s going
to be helpful which relates all of these quantities from the question. That formula says that 𝜀, the
induced potential difference, is equal to negative 𝐿, the self-inductance,
multiplied by 𝛥𝐼, the change in current, divided by 𝛥𝑡, the change in time over
which the current change occurs.
The question tells us that the loop
has a self-inductance of 125 millihenries. So that’s our value for the
quantity 𝐿. We are asked for the time interval
that it takes for the current to increase by 0.25 amperes. So this 0.25 amperes is our value
for the change in current 𝛥𝐼. We are also told that the potential
difference across the loop of wire is 1.2 volts. But actually, our value for the
induced potential difference 𝜀 is going to be negative 1.2 volts.
So why do we have this negative
sign here? Well, if we look at this equation,
we can see that we’ve got a negative sign on the right-hand side. This sign indicates the polarity of
the induced potential difference. What the equation is saying is that
if the current in a loop of wire with a self-inductance of 𝐿 changes by an amount
𝛥𝐼 over a time interval of 𝛥𝑡, then this induces a potential difference 𝜀
across the loop. The magnitude of this potential
difference is equal to 𝐿 multiplied by 𝛥𝐼 over 𝛥𝑡. This negative sign, which we’ve
said indicates the polarity, is telling us that this induced potential difference
tends to generate current that opposes the change in current over time.
So getting back to our values for
the quantities, since we have a positive change in current 𝛥𝐼, then the induced
potential difference, which we know will oppose this, must have a negative
polarity. Another way to reason this is to
notice that the change in time 𝛥𝑡 must have a positive value. Since we know that our values for
𝐿 and 𝛥𝐼 are also positive, then with this negative sign out front, the overall
sign of the right-hand side of the equation must be negative. This means that the left-hand side,
the induced potential difference, must also be negative.
Okay, so in this equation, we know
the values of 𝜀, 𝐿, and 𝛥𝐼. And we’re trying to solve for the
value of 𝛥𝑡. This means that we need to
rearrange the equation to make 𝛥𝑡 the subject. The first step is to multiply both
sides of the equation by the change in time 𝛥𝑡. On the right-hand side, the 𝛥𝑡 in
the numerator cancels with the 𝛥𝑡 in the denominator. Then, we divide both sides of the
equation by the induced potential difference 𝜀. This means that the 𝜀 in the
numerator of the left-hand side cancels with the 𝜀 in the denominator. We end up with an equation that
says 𝛥𝑡 is equal to negative 𝐿 multiplied by 𝛥𝐼 divided by 𝜀.
In order to calculate a change in
time 𝛥𝑡 with units of seconds, we need a self-inductance 𝐿 with units of henries,
a change in current 𝛥𝐼 with units of amperes, and an induced emf 𝜀 with units of
volts. Our value for the induced potential
difference is indeed in units of volts and our change in current is in units of
amperes. However, our value for the
self-inductance of the loop of wire is in units of millihenries rather than
henries. We can recall that 1000
millihenries is equal to one henry, or equivalently one millihenry is equal to one
thousandth of a henry.
So to convert our self-inductance
𝐿 into units of henries, we take its value in units of millihenries and we multiply
it by one over 1000 henries per millihenry. The units of millihenries and per
millihenry cancel each other out. And this leaves us with units of
henries. Then, evaluating the expression, we
find that 𝐿 is equal to 0.125 henries.
Now that we’ve got all of our
quantities in the correct units, we’re ready to go ahead and substitute the values
into this equation. When we do this, we get that 𝛥𝑡
is equal to negative 0.125 henries, that’s our value for 𝐿, multiplied by 0.25
amperes, that’s the change in current 𝛥𝐼, divided by negative 1.2 volts, our value
for 𝜀. We’ve got two negative signs on the
right-hand side of this expression, and so those signs cancel each other out,
leaving us with a positive value.
When we evaluate the right-hand
side of this expression, we find that 𝛥𝑡 is equal to 0.0260416 recurring
seconds. We are told to give our answer to
two decimal places. To this level of precision, our
result for 𝛥𝑡 rounds up to 0.03 seconds. So our answer to the question is
that the amount of time required for the loop of wire to increase the current
through it by 0.25 amperes is 0.03 seconds.
