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Chapter 28: Problem 46

A current of \(2.00 \mathrm{~A}\) is flowing through a 1000 -turn solenoid of length \(L=40.0 \mathrm{~cm} .\) What is the magnitude of the magnetic field inside the solenoid?

Short Answer

Expert verified
Answer: The magnitude of the magnetic field inside the solenoid is approximately \(6.28 × 10^{-4} \, T\).

Step by step solution

01

Identify given variables and constants

We are given the following information: - Current (\(I\)): \(2.00 \, A\) - Number of turns (\(N\)): \(1000\) - Length of the solenoid (\(L\)): \(40.0 \, cm = 0.4 \, m\) - Permeability of free space (\(\mu_0\)): \(4 \pi × 10^{-7} \, Tm/A\)
02

Calculate the number of turns per unit length (n)

To find the number of turns per unit length (n), we'll use the formula \(n = \frac{N}{L}\). Using the given values: $$ n = \frac{1000}{0.4} = 2500 \, turns/m $$
03

Calculate the magnitude of the magnetic field (B)

Now, we can use the formula for the magnetic field inside a solenoid: \(B = \mu_0 n I\). Plug in the values and calculate B: $$ B = (4 \pi × 10^{-7} \, Tm/A) × (2500 \, turns/m) × (2 \, A) \\ B = (4 \pi × 10^{-7} \, Tm/A) × 5000 \, A/m \\ B \approx 6.28 × 10^{-4} \, T $$ So, the magnitude of the magnetic field inside the solenoid is approximately \(6.28 × 10^{-4} \, T\).

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Most popular questions from this chapter

A square loop, with sides of length \(L\), carries current i. Find the magnitude of the magnetic field from the loop at the center of the loop, as a function of \(i\) and \(L\).

Suppose that the magnetic field of the Earth were due to a single current moving in a circle of radius \(2.00 \cdot 10^{3} \mathrm{~km}\) through the Earth's molten core. The strength of the Earth's magnetic field on the surface near a magnetic pole is about \(6.00 \cdot 10^{-5} \mathrm{~T}\). About how large a current would be required to produce such a field?

A horizontally oriented coil of wire of radius \(5.00 \mathrm{~cm}\) and carrying a current, \(i\), is being levitated by the south pole of a vertically oriented bar magnet suspended above the center of the coil. If the magnetic field on all parts of the coil makes an angle \(\theta\) of \(45.0^{\circ}\) with the vertical, determine the magnitude and the direction of the current needed to keep the coil floating in midair. The magnitude of the magnetic field is \(B=0.0100 \mathrm{~T}\), the number of turns in the coil is \(N=10.0\), and the total coil mass is \(10.0 \mathrm{~g}\).

The number of turns in a solenoid is doubled, and its length is halved. How does its magnetic field change? a) it doubles b) it is halved c) it quadruples d) it remains unchanged

The magnetic character of bulk matter is determined largely by electron spin magnetic moments, rather than by orbital dipole moments. (Nuclear contributions are negligible, as the proton's spin magnetic moment is about 658 times smaller than that of the electron.) If the atoms or molecules of a substance have unpaired electron spins, the associated magnetic moments give rise to paramagnetic behavior or to ferromagnetic behavior if the interactions between atoms or molecules are strong enough to align them in domains. If the atoms or molecules have no net unpaired spins, then magnetic perturbations of the electron orbits give rise to diamagnetic behavior. a) Molecular hydrogen gas \(\left(\mathrm{H}_{2}\right)\) is weakly diamagnetic. What does this imply about the spins of the two electrons in the hydrogen molecule? b) What would you expect the magnetic behavior of atomic hydrogen gas (H) to be?
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