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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

In a solenoid in which the wires are wound such that each loop touches the adjacent ones, which of the following will increase the magnetic field inside the magnet? a) making the radius of the loops smaller b) increasing the radius of the wire c) increasing the radius of the solenoid d) decreasing the radius of the wire e) immersion of the solenoid in gasoline

When a magnetic dipole is placed in a magnetic field, it has a natural tendency to minimize its potential energy by aligning itself with the field. If there is sufficient thermal energy present, however, the dipole may rotate so that it is no longer aligned with the field. Using \(k_{\mathrm{B}} T\) as a measure of the thermal energy, where \(k_{\mathrm{B}}\) is Boltzmann's constant and \(T\) is the temperature in kelvins, determine the temperature at which there is sufficient thermal energy to rotate the magnetic dipole associated with a hydrogen atom from an orientation parallel to an applied magnetic field to one that is antiparallel to the applied field. Assume that the strength of the field is \(0.15 \mathrm{~T}\)

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?

What is the magnitude of the magnetic field inside a long, straight tungsten wire of circular cross section with diameter \(2.4 \mathrm{~mm}\) and carrying a current of \(3.5 \mathrm{~A}\), at a distance of \(0.60 \mathrm{~mm}\) from its central axis?

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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