Question

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

Answer: The ratio of the magnitudes of the magnetic fields in the interiors of solenoid A to solenoid B is \(\frac{4}{3}\).

Step-by-step solution

Calculate the number of turns per unit length for both solenoids

Since we have the properties of solenoids A and B, we can calculate the number of turns per unit length (\(n\)) for both solenoids. Solenoid A has four times the number of turns of solenoid B, and three times the length, so the turns per unit length for solenoid A is: n_A = \(\frac{4N}{3L}\), where \(N\) is the number of turns of solenoid B and \(L\) is the length of solenoid B. Solenoid B has \(N\) turns and length \(L\), so the turns per unit length for solenoid B is: n_B = \(\frac{N}{L}\)

Calculate the magnetic fields for both solenoids

Now, we need to calculate the magnetic fields for both solenoids using the formula \(B = \mu_0 n I\). Because the currents flowing through both solenoids have equal magnitudes, we have: \(B_A = \mu_0 n_A I\) and \(B_B = \mu_0 n_B I\)

Find the ratio of the magnitudes of the magnetic fields

To find the ratio of the magnitudes of the magnetic fields, we divide \(B_A\) by \(B_B\): \(\frac{B_A}{B_B} = \frac{\mu_0 n_A I}{\mu_0 n_B I}\) The current \(I\) and permeability of free space \(\mu_0\) are equal in both solenoids, so they cancel out: \(\frac{B_A}{B_B} = \frac{n_A}{n_B}\) Now we can substitute the values for \(n_A\) and \(n_B\) we found in step 1: \(\frac{B_A}{B_B} = \frac{\frac{4N}{3L}}{\frac{N}{L}}\) Finally, simplify the expression by cancelling out the \(N\) and \(L\) terms and obtain the answer: \(\frac{B_A}{B_B} = \frac{4N}{3N} = \frac{4}{3}\) This means that the magnitude of the magnetic field in the interior of solenoid A is \(\frac{4}{3}\) times the magnitude of the magnetic field in the interior of solenoid B.

Key concepts

Solenoid Magnetic Field Calculation

Understanding how to calculate the magnetic field of a solenoid is crucial for students studying electromagnetism. A solenoid is a type of electromagnet comprised of a coil of wire with an electric current passing through it. When the electric current flows through the wire, it generates a magnetic field within the solenoid that is relatively uniform and aligned along the axis of the coil.

The magnetic field inside a long solenoid can be calculated using the formula:
\[B = \f{\f0}{\f2}{mu_0} n I\]
where:

• \(B\) is the magnetic field,
• \(\f{\f1}{\f2}{mu_0}\) is the magnetic constant or permeability of free space (\(4\fix5}13\times 10^{-7} N/A^2\)),
• \(n\) is the number of turns per unit length of the solenoid, and
• \(I\) is the current passing through the solenoid.

By manipulating this formula, it's possible to compare the strength of magnetic fields between two solenoids, taking into account the differences in their physical characteristics and the currents flowing through them as exemplified in the textbook solution.

Magnetic Field Ratio

The magnetic field ratio between two solenoids is a simple yet informative measure that compares their magnetic field strengths. This ratio is essential to understand because it allows us to determine how changes in a solenoid’s characteristics—like the number of turns or its dimensions—can affect the magnetic field it produces.

In the given exercise, we computed the magnetic field ratio by using the formula for magnetic field strength and comparing the properties of two solenoids. The magnetic field ratio is particularly insightful because it shows the direct proportionality to the turns per unit length and the current when the solenoids have the same type of wire and are in a similar environment. In the exercise, we found that the ratio of the magnetic fields is independent of the solenoid dimensions when the current and permeability of free space are constants.

Turns Per Unit Length

The concept of turns per unit length in a solenoid is integral to understanding how tightly the wire is wound around the solenoid's core. It is defined as the number of turns of the wire divided by the length of the solenoid. Mathematically, it is represented by
\[n = \frac{N}{L}\]
where \(n\) stands for turns per unit length, \(N\) represents the total number of turns of the coil, and \(L\) is the length of the solenoid.

This value is crucial in the calculation of the magnetic field of a solenoid because a higher number of turns per unit length will generally result in a stronger magnetic field, assuming the current is constant. In the given exercise solution, calculating the turns per unit length for both solenoids allowed us to directly compare their magnetic fields.

Ampere's Law for Solenoids

Ampere's Law is a foundational principle in magnetostatics, which relates the integrated magnetic field around a closed loop to the electric current passing through the loop. In the context of solenoids, Ampere’s Law can be simplistically applied when the solenoid is long and the field lines run parallel to its length. This ideal condition lets us assume a uniform magnetic field inside the solenoid.

In practical terms, for a solenoid with a high number of turns that is sufficiently long, Ampere's Law helps us derive the magnetic field formula \(B = \f{\f1}{\f2}{mu_0} n I\). This approach assumes the magnetic field outside the solenoid is negligible compared to the inside and is a manifestation of how Ampere's Law can predict magnetic field behavior in a solenoid provided we know the current and turns per unit length.