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Chapter 19: Problem 74

A solenoid has 4850 turns per meter and radius \(3.3 \mathrm{cm}\) The magnetic field inside has magnitude 0.24 T. What is the current in the solenoid?

Short Answer

Expert verified
Answer: The current flowing through the solenoid is approximately 3.11 A.

Step by step solution

01

Write down the given values and the formula for the magnetic field inside a solenoid.

The given values are number of turns per meter (\(n = 4850 \, turns/m\)), radius (\(r = 3.3 \, cm\)), and the magnetic field inside the solenoid (\(B = 0.24 \, T\)). The permeability of free space (\(\mu_0 = 4\pi \times 10^{-7} \, Tm/A\)) is a constant value. We will use Amp猫re's law: $$ B = \mu_0 \cdot n \cdot I $$
02

Solve the formula for current I.

We want to find the current I in the solenoid. Therefore, rearrange the formula to get I by itself: $$ I = \frac{B}{\mu_0 \cdot n} $$
03

Plug in the given values and solve for I.

Now, we can substitute the given values into the formula and calculate the current I: $$ I = \frac{0.24 T}{(4\pi \times 10^{-7} \, Tm/A)(4850 \, turns/m)} $$ After plugging in the numbers and doing the calculations, we get: $$ I \approx 3.11 A $$ So the current in the solenoid is approximately 3.11 A.

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Most popular questions from this chapter

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In a simple model, the electron in a hydrogen atom orbits the proton at a radius of \(53 \mathrm{pm}\) and at a constant speed of $2.2 \times 10^{6} \mathrm{m} / \mathrm{s} .$ The orbital motion of the electron gives it an orbital magnetic dipole moment. (a) What is the current \(I\) in this current loop? [Hint: How long does it take the electron to make one revolution?] (b) What is the orbital dipole moment IA? (c) Compare the orbital dipole moment with the intrinsic magnetic dipole moment of the electron $\left(9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}\right)$
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