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Chapter 37: Problem 55

The critical field in a niobium-titanium superconductor is \(15 \mathrm{T}\) What current in a 5000 -turn solenoid \(75 \mathrm{cm}\) long will produce a field of this strength?

Short Answer

Expert verified
The current needed to produce a magnetic field of 15 T in a 5000-turn solenoid that is 75 cm long is \(7.16 *10^{4} \, \text{A}\)

Step by step solution

01

Write down the given parameters

B = \(15 \, \text{T}\)\nN = \(5000 \, \text{turns}\)\nL = \(75 \, \text{cm} = 0.75 \, \text{m}\)\n\(\mu_0 = 4\pi * 10^{-7} \, \text{Tm/A}\)
02

Identify the formula and rearrange it to solve for the current

The formula for the magnetic field in a solenoid is \(B= \mu_0 * \frac{N*I}{L}\).\nRearrange this formula to solve for the current, \(I\):\n\(I = \frac{B * L}{\mu_0 * N}\)
03

Substitute the given values and solve

Substitute the values of the magnetic field \(B\), length of the solenoid \(L\), number of turns \(N\), and the permeability of the free space \(\mu_0\) into the formula to find the current \(I\):\nI = \(\frac{15 * 0.75}{4\pi * 10^{-7} * 5000} = 7.16 *10^{4} \, \text{A}\).

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