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Chapter 20: Problem 46

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 桅 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 桅 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 桅 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 桅 through one winding using the formula: \(\Phi = B \times A\) Substitute the previously calculated values for B and A to find the magnetic flux: \(\Phi = B \times A\)

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Most popular questions from this chapter

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?
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 \(I(t) .]\)
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