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Chapter 20: Problem 76

A solenoid is \(8.5 \mathrm{cm}\) long, \(1.6 \mathrm{cm}\) in diameter, and has 350 turns. When the current through the solenoid is \(65 \mathrm{mA},\) what is the magnetic flux through one turn of the solenoid?

Short Answer

Expert verified
Solution: The magnetic flux through one turn of the solenoid is approximately \(7.00 \times 10^{-6} \, Wb\).

Step by step solution

01

Find the number of turns per unit length

To find the number of turns per unit length (\(n\)), we need to divide the total number of turns (350) by the length of the solenoid (8.5 cm). It's important to convert the length to meters. \(n = \dfrac{350}{0.085} \approx 4117.65 \, turns/m\)
02

Calculate the magnetic field inside the solenoid

To calculate the magnetic field inside the solenoid, we use the formula \(B = \mu_0 * n * I\). The permeability of free space, \(\mu_0\), is a constant equal to \(4\pi 脳 10^{-7} \, \mathrm{T\cdot m/A}\). The current (\(I\)) is given as 65 mA, but we need to convert it to amperes (A). \(I = 65 \times 10^{-3} \, A\) Now, we can plug in the values to find the magnetic field: \(B = (4\pi 脳 10^{-7} \, \mathrm{T\cdot m/A}) * 4117.65 \, turns/m * (65 \times 10^{-3} \, A) \approx 0.0348 \, T\)
03

Calculate the area of a single turn

To calculate the area of a single turn, we can use the formula for the area of a circle, \(A = \pi r^2\), where \(r\) is the radius. The diameter of the solenoid is given as 1.6 cm, so the radius is half of that, which is 0.8 cm. We need to convert the radius to meters before plugging it into the formula. \(r = 0.008 \, m\) Now we can find the area: \(A = \pi (0.008)^2 = 2.01 \times 10^{-4} \, m^2\)
04

Calculate the magnetic flux through one turn of the solenoid

Finally, we can calculate the magnetic flux using the formula \(\Phi = B * A\). We have the magnetic field (\(B = 0.0348 \, T\)) and the area of a single turn (\(A = 2.01 \times 10^{-4} \, m^2\)). \(\Phi = 0.0348 \, T * 2.01 \times 10^{-4} \, m^2 \approx 7.00 \times 10^{-6} \, Wb\) So, the magnetic flux through one turn of the solenoid is approximately \(7.00 \times 10^{-6} \, Wb\).

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Most popular questions from this chapter

A coil of wire is connected to an ideal \(6.00-\mathrm{V}\) battery at \(t=0 .\) At \(t=10.0 \mathrm{ms},\) the current in the coil is \(204 \mathrm{mA}\) One minute later, the current is 273 mA. Find the resistance and inductance of the coil. [Hint: Sketch \(I(t) .]\)
An ideal solenoid ( \(N_{1}\) turns, length \(L_{1},\) radius \(r_{1}\) ) is placed inside another ideal solenoid ( \(N_{2}\) turns, length \(L_{2}>L_{1}\), radius \(r_{2}>r_{1}\) ) such that the axes of the two coincide. (a) What is the mutual inductance? (b) If the current in the outer solenoid is changing at a rate \(\Delta I_{2} / \Delta t,\) what is the magnitude of the induced emf in the inner solenoid?
The largest constant magnetic field achieved in the laboratory is about $40 \mathrm{T}$. (a) What is the magnetic energy density due to this field? (b) What magnitude electric field would have an equal energy density?
The armature of an ac generator is a rectangular coil \(2.0 \mathrm{cm}\) by \(6.0 \mathrm{cm}\) with 80 turns. It is immersed in a uniform magnetic field of magnitude 0.45 T. If the amplitude of the emf in the coil is $17.0 \mathrm{V},$ at what angular speed is the armature rotating?
Suppose you wanted to use Earth's magnetic field to make an ac generator at a location where the magnitude of the field is \(0.50 \mathrm{mT}\). Your coil has 1000.0 turns and a radius of \(5.0 \mathrm{cm} .\) At what angular velocity would you have to rotate it in order to generate an emf of amplitude $1.0 \mathrm{V} ?$
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