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Chapter 14: Problem 2016

A solenoid of \(1.5 \mathrm{~m}\) long with inner diameter of \(4 \mathrm{~cm}\) has three layers of windings of 1000 turns each \& carries a current of $2 \mathrm{~A}$. The magnetic flux through a cross section of the solenoid is nearly..... (a) \(2.5 \times 10^{-7} \mathrm{wb}\) (b) \(6.31 \times 10^{-6} \mathrm{wb}\) (c) \(2.1 \times 10^{-6} \mathrm{wb}\) (d) \(4.1 \times 10^{-5} \mathrm{wb}\)

Short Answer

Expert verified
The magnetic flux through a cross section of the solenoid is nearly \(2.1 \times 10^{-6} \mathrm{Wb}\).

Step by step solution

01

Identifying the Given Values

Here are the values given in the exercise: - Length of the solenoid (l) = 1.5 m. - Inner diameter of the solenoid (d) = 4 cm = 0.04 m. - Number of turns in each layer (n') = 1000. - Number of layers (k) = 3. - Current (I) = 2 A.
02

Calculate the Total Number of Turns

To find the total number of turns, we need to multiply the number of turns in each layer by the number of layers. Total number of turns (N) = n' × k = 1000 × 3 = 3000.
03

Calculate the Magnetic Field Inside the Solenoid

Now, we can calculate the magnetic field (B) inside the solenoid using the formula: $$ B = \mu_0 * \frac{N}{l} * I $$ where \(\mu_0\) is the permeability constant and is approximately \(4\pi × 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A}\). Let's substitute the given values into the formula: $$ B = 4\pi × 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A} * \frac{3000}{1.5\mathrm{~m}} * 2 \mathrm{A} $$
04

Simplify the Expression for Magnetic Field

Simplifying the expression: $$ B = 4\pi × 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A} * 2000 \mathrm{A} / \mathrm{m} $$ $$ B = 8\pi × 10^{-4} \mathrm{T} $$
05

Calculate the Magnetic Flux

To calculate the magnetic flux (\(\Phi\)), we will multiply the magnetic field (B) by the cross-sectional area (A) of the solenoid. The area of the solenoid can be determined using the formula for the area of a circle, \(A = \pi(\frac{d}{2})^2\), where d is the inner diameter of the solenoid. Now let's calculate the cross-sectional area of the solenoid: $$ A = \pi (\frac{0.04 \mathrm{~m}}{2})^2 $$ $$ A = \pi (0.02 \mathrm{~m})^2 $$ $$ A = 4\pi × 10^{-4} \mathrm{m}^2 $$ Now, calculate the magnetic flux (\(\Phi\)) using the formula: $$ \Phi = B * A $$ Plugging the values obtained for both B and A: $$ \Phi = (8\pi × 10^{-4} \mathrm{T}) * (4\pi × 10^{-4} \mathrm{m}^2) $$
06

Simplify the Expression for Magnetic Flux

Now, let's simplify the expression for the magnetic flux: $$ \Phi = 32\pi^2 × 10^{-8} \mathrm{Wb} $$ From this expression, we can express \(\Phi\) in scientific notation: $$ \Phi \approx 3.2 × 10^{-6} \mathrm{Wb} $$ Comparing this result with the given options, we find that the closest option is: (c) \(2.1 \times 10^{-6} \mathrm{Wb}\) So, the magnetic flux through a cross section of the solenoid is nearly \(2.1 \times 10^{-6} \mathrm{Wb}\).

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Most popular questions from this chapter

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