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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 \(4\pi × 10^{-7} \, \mathrm{T\cdot m/A}\). 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: \(B = (4\pi × 10^{-7} \, \mathrm{T\cdot m/A}) * 4117.65 \, turns/m * (65 \times 10^{-3} \, A) \approx 0.0348 \, T\)
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 100 -turn coil with a radius of \(10.0 \mathrm{cm}\) is mounted so the coil's axis can be oriented in any horizontal direction. Initially the axis is oriented so the magnetic flux from Earth's field is maximized. If the coil's axis is rotated through \(90.0^{\circ}\) in \(0.080 \mathrm{s},\) an emf of $0.687 \mathrm{mV}$ is induced in the coil. (a) What is the magnitude of the horizontal component of Earth's magnetic field at this location? (b) If the coil is moved to another place on Earth and the measurement is repeated, will the result be the same?
A flip coil is a device used to measure a magnetic field. A coil of radius $r, N\( turns, and electrical resistance \)R$ is initially perpendicular to a magnetic field of magnitude B. The coil is connected to a special kind of galvanometer that measures the total charge \(Q\) that flows through it. To measure the field, the flip coil is rapidly flipped upside down. (a) What is the change in magnetic flux through the coil in one flip? (b) If the time interval during which the coil is flipped is \(\Delta t,\) what is the average induced emf in the coil? (c) What is the average current that flows through the galvanometer? (d) What is the total charge \(Q\) in terms of \(r, N, R,\) and \(B ?\)
The windings of an electromagnet have inductance \(L=8.0 \mathrm{H}\) and resistance \(R=2.0 \Omega . \quad A \quad 100.0-V \quad d c\) power supply is connected to the windings by closing switch \(S_{2} .\) (a) What is the current in the windings? (b) The electromagnet is to be shut off. Before disconnecting the power supply by opening switch \(S_{2},\) a shunt resistor with resistance \(20.0 \Omega\) is connected in parallel across the windings. Why is the shunt resistor needed? Why must it be connected before the power supply is disconnected? (c) What is the maximum power dissipated in the shunt resistor? The shunt resistor must be chosen so that it can handle at least this much power without damage. (d) When the power supply is disconnected by opening switch \(S_{2},\) how long does it take for the current in the windings to drop to \(0.10 \mathrm{A} ?\) (e) Would a larger shunt resistor dissipate the energy stored in the electromagnet faster? Explain.
Calculate the equivalent inductance \(L_{\mathrm{eq}}\) of two ideal inductors, \(L_{1}\) and \(L_{2},\) connected in series in a circuit. Assume that their mutual inductance is negligible. [Hint: Imagine replacing the two inductors with a single equivalent inductor \(L_{\mathrm{cq}} .\) How is the emf in the series equivalent related to the emfs in the two inductors? What about the currents?]
A coil has an inductance of \(0.15 \mathrm{H}\) and a resistance of 33 \(\Omega\). The coil is connected to a 6.0 - \(V\) ideal battery. When the current reaches half its maximum value: (a) At what rate is magnetic energy being stored in the inductor? (b) At what rate is energy being dissipated? (c) What is the total power that the battery supplies?
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