If you want to construct an electromagnet by running a current of 3.0 A through a solenoid with 500 windings and length \(3.5 \mathrm{~cm}\) and you want the magnetic field inside the solenoid to have magnitude \(B=2.96 \mathrm{~T}\), you can insert a ferrite core into the solenoid. What value of the relative magnetic permeability should this ferrite core have in order to make this work?

Short Answer

Expert verified
Answer: The relative magnetic permeability of the ferrite core needed is approximately 699.43.

Step by step solution

01

Known values and constants

We are given the following information: - Current, \(I = 3.0 \mathrm{~A}\) - Number of windings, \(N = 500\) - Solenoid length, \(l = 3.5 \mathrm{~cm} = 0.035 \mathrm{~m}\) - Magnetic field, \(B = 2.96 \mathrm{~T}\) - Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \mathrm{~Tm/A}\)
02

Calculate the number of turns per unit length

We can calculate the number of turns per unit length (\(n\)) by dividing the total number of windings by the length of the solenoid: \(n = \frac{N}{l} = \frac{500}{0.035 \mathrm{~m}} = 14285.71 \mathrm{~turns/m}\)
03

Rearrange the formula for the magnetic field

We rearrange the formula for the magnetic field inside a solenoid with a core to solve for \(\mu_r\): \(\mu_r = \frac{B}{\mu_0 n I}\)
04

Plug in the values and solve for the relative magnetic permeability

Now we can plug in the values and solve for \(\mu_r\): \(\mu_r = \frac{2.96 \mathrm{~T}}{(4\pi \times 10^{-7} \mathrm{~Tm/A})(14285.71 \mathrm{~turns/m})(3.0 \mathrm{~A})} = 699.43\) The relative magnetic permeability of the ferrite core should be approximately 699.43 to create the desired magnetic field.

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Most popular questions from this chapter

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