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A solenoid is formed of a length of wire that carries a constant current πΌ.
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The solenoid has 430 turns of wire per meter.
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The magnetic field at the center of the solenoid is measured to be 3.2 times 10 to the negative three tesla.
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Calculate the current πΌ in amperes.
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Give your answer to one decimal place.
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Use π naught is equal to four π times 10 to the negative seven tesla meters per ampere.
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In this question, weβre being asked about a solenoid, which is a wire thatβs formed into a series of equally spaced loops or turns like weβve illustrated here.
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Weβre told that this wire carries a constant current of πΌ, and this current is what weβre being asked to find.
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As a result of this current in the wire, thereβs a magnetic field inside of the solenoid.
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Weβre told that the strength of this magnetic field, which weβve labeled as π΅, is measured at the center of the solenoid to be equal to 3.2 times 10 to the negative three tesla.
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We can recall that for a solenoid that has a total length of πΏ and consists of π turns of wire which carry a current of πΌ, the magnetic field π΅ inside of that solenoid is equal to π naught multiplied by capital π multiplied by πΌ divided by πΏ, where π naught is a constant known as the permeability of free space.
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This equation here relates the magnetic field π΅ inside of the solenoid, which we know the value of, to the current πΌ in the wire, which is what we want to work out.
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However, we donβt know the total number of turns of wire in this solenoid capital π, and we also donβt know the solenoidβs length πΏ.
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What we are told is that this solenoid has 430 turns of wire per meter.
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If we define a quantity lowercase π as the number of turns of wire per unit length of the solenoid, then we have for this solenoid that lowercase π is equal to 430 meters to the negative one because it has 430 turns of wire per meter of length.
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We can see that the number of turns of wire per unit length, lowercase π, must be equal to the total number of turns capital π divided by the total length πΏ.
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Since in this equation here we donβt know the value for either capital π or for πΏ but we do know the value of lowercase π and that lowercase π is equal to capital π divided by πΏ, letβs use this relationship in order to replace capital π divided by πΏ in this equation for the magnetic field strength by lowercase π, the number of turns per unit length.
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When we do this, we find that π΅, the strength of the magnetic field inside of the solenoid, is equal to π naught, the permeability of free space, multiplied by lowercase π, the number of turns of wire per unit length, multiplied by the current πΌ in the wire.
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Now, for this solenoid, we know the values of both π΅ and lowercase π.
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And weβre also given a value for this constant π naught.
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That means that the only unknown quantity in the equation is the current πΌ that weβre trying to work out.
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In order to work out this current, we need to make πΌ the subject of the equation.
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To do this, weβll divide both sides of the equation by π naught and π so that on the right-hand side the π naughts and πβs cancel from the numerator and denominator.
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If we then swap over the left- and right-hand sides of the equation, we have that the current πΌ is equal to π΅ divided by π naught and lowercase π.
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Weβre now ready to go ahead and substitute in our values for the strength of the magnetic field π΅, the number of turns of wire per unit length lowercase π, and the permeability of free space π naught.
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When we do this, we get that πΌ is equal to 3.2 times 10 to the negative three tesla divided by four π times 10 to the negative seven tesla meters per ampere and 430 meters to the negative one.
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Looking at the units on the right-hand side of the equation, we can see that in the denominator the meters and the meters to the negative one cancel each other out.
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Then, we can also see that the teslas cancel from the numerator and denominator, and this just leaves us with units of one over amperes in the denominator of the fraction.
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The overall units weβll get for our current πΌ then will be units of one divided by one over amperes.
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This is simply equal to units of amperes, which is the current unit that the question asks us to use.
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Evaluating the expression, we calculate a current πΌ of 5.922 et cetera amperes.
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We should notice that weβre asked to give this answer to one decimal place.
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Rounding to one decimal place of precision, the result becomes 5.9 amperes.
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Our answer then is that the current in the wire is equal to 5.9 amperes.