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