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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 long, straight wire has a 10.0 - A current flowing in the positive \(x\) -direction, as shown in the figure. Close to the wire is a square loop of copper wire that carries a 2.00 - A cur- rent in the direction shown. The near side of the loop is \(d=0.50 \mathrm{~m}\) away from the wire. The length of each side of the square is \(a=1.00 \mathrm{~m}\). a) Find the net force between the two current-carrying objects. b) Find the net torque on the loop.

Two long, parallel wires separated by a distance, \(d\), carry currents in opposite directions. If the left-hand wire carries a current \(i / 2,\) and the right-hand wire carries a current \(i\), determine where the magnetic field is zero.

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28.39 The figure shows a cross section across the diameter of a long, solid, cylindrical conductor. The radius of the cylinder is \(R=10.0 \mathrm{~cm} .\) A current of \(1.35 \mathrm{~A}\) is uniformly distributed through the conductor and is flowing out of the page. Calculate the direction and the magnitude of the magnetic field at positions \(r_{\mathrm{a}}=0.0 \mathrm{~cm}\) \(r_{\mathrm{b}}=4.00 \mathrm{~cm}, r_{\mathrm{c}}=10.00 \mathrm{~cm}\) and \(r_{d}=16.0 \mathrm{~cm} .\)

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