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Chapter 28: Problem 47

Solenoid A has twice the diameter, three times the length, and four times the number of turns of solenoid B. The two solenoids have currents of equal magnitudes flowing through them. Find the ratio of the magnitude of the magnetic field in the interior of solenoid \(A\) to that of solenoid \(B\)

Short Answer

Expert verified
Answer: The ratio of the magnetic field inside solenoid A to the magnetic field inside solenoid B is 4:3.

Step by step solution

01

Recall the formula for the magnetic field inside a solenoid

The magnetic field B inside a solenoid is given by the formula: B = μ₀ * n * I where μ₀ is the permeability of free space (μ₀ ≈ 4π × 10^(-7) Tm/A), n is the number of turns per unit length (turns/m), and I is the current (in A).
02

Find the number of turns per unit length of solenoids A and B

We are given that solenoid A has 4 times the number of turns as solenoid B. Let n_B be the number of turns per unit length of solenoid B, so the number of turns per unit length for solenoid A will be 4n_B. We also know that solenoid A is three times longer than solenoid B, so the length of solenoid A is 3 times the length of solenoid B. Let l_B be the length of solenoid B, then the length of solenoid A will be 3l_B. Now we can find the number of turns per unit length for solenoids A and B: n_A = (4n_B) / (3l_B) = (4/3)n_B n_B = n_B
03

Find the ratio of magnetic fields

Now we can use the formula for magnetic fields inside solenoids to find the ratio of magnetic fields for solenoids A and B: B_A / B_B = (μ₀ * n_A * I) / (μ₀ * n_B * I) Since the currents in both solenoids have equal magnitudes, the current factor in the above equation cancels out: B_A / B_B = n_A / n_B Now we can substitute the values we found for n_A and n_B in step 2: B_A / B_B = ((4/3)n_B) / n_B The n_B factors cancel out, leaving us with: B_A / B_B = 4/3 So, the ratio of the magnitude of the magnetic field in the interior of solenoid A to that of solenoid B is 4:3.

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Most popular questions from this chapter

Many electrical applications use twisted-pair cables in which the ground and signal wires spiral about each other. Why?

Two identical coaxial coils of wire of radius \(20.0 \mathrm{~cm}\) are directly on top of each other, separated by a 2.00 -mm gap. The lower coil is on a flat table and has a current \(i\) in the clockwise direction; the upper coil carries an identical current and has a mass of \(0.0500 \mathrm{~kg} .\) Determine the magnitude and the direction that the current in the upper coil has to have to keep the coil levitated at its current height.

A circular wire of radius \(5.0 \mathrm{~cm}\) has a current of \(3.0 \mathrm{~A}\) flowing in it. The wire is placed in a uniform magnetic field of \(5.0 \mathrm{mT}.\) a) Determine the maximum torque on the wire. b) Determine the range of the magnetic potential energy of the wire.

Two long, straight wires are parallel to each other. The wires carry currents of different magnitudes. If the amount of current flowing in each wire is doubled, the magnitude of the force between the wires will be a) twice the magnitude of the original force. b) four times the magnitude of the original force. c) the same as the magnitude of the original force. d) half of the magnitude of the original force.

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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