WEBVTT
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A 1.0 times 10 to the two volt potential difference is applied across a 10.0-metre length of wire with a diameter of 4.621 millimetres.
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The magnitude of the current density produced is 2.0 times 10 to the eighth amps per metre squared.
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What is the resistivity of the wire?
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We can call the potential difference applied across the wire 1.0 times 10 to the two volts π£.
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The length of the wire 10.0 metres, we can call capital πΏ.
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Weβre told the wire has a diameter of 4.621 millimetres, which weβll called π.
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And the current density running through the wire is given as 2.0 times 10 to the eighth amps per metre squared.
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Weβll call that π½.
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We want to know the resistivity of the wire, which weβll call π.
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We can start off by recalling a relationship between resistivity and resistance.
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The electrical resistance π
of a wire is equal to the resistivity π of the material the wire is made of times the length of the wire divided by the cross-sectional area of the wire π΄.
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Rearranging this equation to solve for π, we see itβs equal to the resistance of the wire multiplied by its cross-sectional area divided by its length πΏ.
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πΏ is given to us in the problem statement.
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And π΄ we can solve for using the diameter of the wire π.
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Assuming the wire has a circular cross section, we recall that the area of the circle is π times its radius squared.
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In our case, π΄ equals π times π over two quantity squared or π over four times 4.621 times 10 to the negative third metres squared.
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Knowing the length of the wire πΏ and being able to calculate its cross-sectional area π΄, we now want to solve for its resistance π
.
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To do that, we can recall Ohmβs law, which says that the voltage π£ across a circuit is equal to the current in the circuit multiplied by its resistance π
.
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π£ being equal to πΌ times π
implies that π
is equal to π£ divided by πΌ.
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Weβre told the potential difference across the circuit in the problem statement, but we donβt know the current involved πΌ.
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However, we can recall that current density π½ is equal to πΌ divided by cross-sectional area π΄.
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π½ being equal to πΌ divided by π΄ implies that πΌ is equal to π½ times π΄.
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π½ is given to us in the problem statement and π΄ is a known value.
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When we plug in for these values to solve for πΌ and enter them on our calculator, we find that the current is approximately 3354 amps.
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Knowing that current, we can now return to our Ohmβs law equation to solve for the resistance π
.
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π
is equal to π£ over πΌ or 1.0 times 10 to the two volts divided by 3354 amps.
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Calculating this fraction, we find that π
is approximately 0.0298 ohms.
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We are now ready to return to our equation for resistivity in terms of resistance π
.
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Weβve calculated π
and π΄ and our given πΏ.
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So weβre ready to plug in and solve for π.
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When we do, we find that π is equal to two significant figures to 5.0 times 10 to the negative eighth ohm metres.
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Thatβs the resistivity of this wire.