A solenoid of length \(2.8 \mathrm{cm}\) and diameter \(0.75 \mathrm{cm}\) is wound with 160 turns per cm. When the current through the solenoid is $0.20 \mathrm{A},$ what is the magnetic flux through one of the windings of the solenoid?

Short Answer

Expert verified
In this problem, we were asked to calculate the magnetic flux Φ through one winding of a solenoid. We followed these steps to find the solution: 1. We calculated the total number of turns N by multiplying the turns per cm by the length of the solenoid (160 turns/cm x 2.8 cm). 2. We calculated the magnetic field B inside the solenoid using the formula \(B = \mu n I\), where \(\mu\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(I\) is the current through the solenoid. 3. We calculated the area A of a single winding loop by finding the radius (half of the diameter) of the solenoid and using the formula for the area of a circle, \(A = \pi r^2\). 4. Finally, we calculated the magnetic flux Φ through one winding using the formula \(\Phi = B \times A\) and found the final answer by substituting the calculated values for B and A.

Step by step solution

01

Calculate the total number of turns N

To calculate the total number of turns N, we need to multiply the turns per cm by the length of the solenoid. In this case, 160 turns per cm and the length is 2.8 cm. \(N = (\)turns per cm\() \times (\)length in cm\() = 160 \times 2.8\)
02

Calculate the magnetic field B inside the solenoid

To calculate the magnetic field B inside the solenoid, we use the formula for the magnetic field of a solenoid: \(B = \mu n I\) where \(\mu = 4\pi \times 10^{-7} \mathrm{T \cdot m/A}\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(I\) is the current through the solenoid. First, we need to find \(n\). Since there are 160 turns per cm, we can convert that to turns per meter: \(n = 160 \times 10^{2} \ \mathrm{turns/m}\). Now we can plug in the values to find the magnetic field B: \(B = (4\pi \times 10^{-7} \mathrm{T \cdot m/A}) \times (160 \times 10^{2} \ \mathrm{turns/m}) \times (0.20 \mathrm{A})\)
03

Calculate the area A of a single winding loop

To find the area A of a single winding loop, we need to know the diameter of the solenoid. In this case, the diameter is 0.75 cm, so the radius is: \(r = \frac{0.75 \ \mathrm{cm}}{2} = 0.375 \ \mathrm{cm}\), which can be converted to meters: \(r = 0.00375 \ \mathrm{m}\) Now, we can find the area A using the formula for the area of a circle: \(A = \pi r^2 = \pi (0.00375 \ \mathrm{m})^2\)
04

Calculate the magnetic flux Φ through one winding

Now that we have the magnetic field B and the area A of a single winding loop, we can calculate the magnetic flux Φ through one winding using the formula: \(\Phi = B \times A\) Substitute the previously calculated values for B and A to find the magnetic flux: \(\Phi = B \times A\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Section \(20.9,\) in order to find the energy stored in an inductor, we assumed that the current was increased from zero at a constant rate. In this problem, you will prove that the energy stored in an inductor is \(U_{\mathrm{L}}=\frac{1}{2} L I^{2}-\) that is, it only depends on the current \(I\) and not on the previous time dependence of the current. (a) If the current in the inductor increases from \(i\) to \(i+\Delta i\) in a very short time \(\Delta t,\) show that the energy added to the inductor is $$\Delta U=L i \Delta i$$ [Hint: Start with \(\Delta U=P \Delta t .]\) (b) Show that, on a graph of \(L i\) versus \(i,\) for any small current interval \(\Delta i,\) the energy added to the inductor can be interpreted as the area under the graph for that interval.(c) Now show that the energy stored in the inductor when a current \(I\) flows is \(U=\frac{1}{2} L I^{2}\)
In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.
A uniform magnetic field of magnitude 0.29 T makes an angle of \(13^{\circ}\) with the plane of a circular loop of wire. The loop has radius $1.85 \mathrm{cm} .$ What is the magnetic flux through the loop?
The magnetic field between the poles of an electromagnet is \(2.6 \mathrm{T} .\) A coil of wire is placed in this region so that the field is parallel to the axis of the coil. The coil has electrical resistance \(25 \Omega,\) radius $1.8 \mathrm{cm},\( and length \)12.0 \mathrm{cm} .$ When the current supply to the electromagnet is shut off, the total charge that flows through the coil is \(9.0 \mathrm{mC} .\) How many turns are there in the coil?
When the emf for the primary of a transformer is of amplitude $5.00 \mathrm{V},\( the secondary emf is \)10.0 \mathrm{V}$ in amplitude. What is the transformer turns ratio \(\left(N_{2} / N_{1}\right) ?\)
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free