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Chapter 26: Problem 41

A superconducting solenoid has 3300 turns per meter and carries 4.1 kA. Find the magnetic field strength in the solenoid.

Short Answer

Expert verified
The magnetic field strength of the superconducting solenoid is approximately 5.38 T.

Step by step solution

01

Identify Given Values

For this problem, the given values are as follows: the number of turns per meter (n) is 3300, and the current (I) is 4.1 kA.
02

Convert the Current Value

The current value is given in kiloamperes (kA). We need to convert this to amperes (A) to use in our formula. Thus, \( I = 4.1 kA = 4100 A \).
03

Identify the Permeability of Free Space Value

The permeability of free space, \( \mu_0 \), is a known constant. Its value is \( 4\pi x 10^{-7} \, Tm/A \)
04

Apply Amp猫re's Circuital Law

Substitute the given values into Amp猫re's circuital law \( B = \mu_0 * n * I \), where \( B \) is the magnetic field strength, \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per meter, and \( I \) is the current. Therefore, the magnetic field strength would be \( B = 4\pi x 10^{-7} Tm/A * 3300 turns/m * 4100 A \).
05

Calculate the Magnetic Field Strength

Now that all the values are substituted into the formula, calculate the magnetic field to get the final answer.

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Most popular questions from this chapter

A proton moving with velocity \(\vec{v}_{1}=3.6 \times 10^{4} \hat{\jmath} \mathrm{m} / \mathrm{s}\) experiences a magnetic force of \(7.4 \times 10^{-16} \hat{\imath} \mathrm{N} .\) A second proton moving on the \(x\) -axis experiences a magnetic force of \(2.8 \times 10^{-16} \hat{\jmath} \mathrm{N}\). Find the magnitude and direction of the magnetic field (assumed uniform), and the velocity of the second proton.

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