Question

A long solenoid with length \(3.00 \mathrm{~m}\) and \(n=290 .\) turns \(/ \mathrm{m}\) carries a current of \(3.00 \mathrm{~A}\). It stores \(2.80 \mathrm{~J}\) of energy. What is the cross-sectional area of the solenoid?

Short answer

Question: Calculate the cross-sectional area of the solenoid given the following parameters: Length of the solenoid is 3.00 m, the number of turns per meter is 290, the current is 3.00 A, and the stored energy is 2.80 J. Answer: The cross-sectional area of the solenoid is approximately \(4.70 \times 10^{-4} \ \mathrm{m^2}\).

Step-by-step solution

Identify the given values

From the problem, we are given the following values: Length of the solenoid \(l= 3.00 \ \mathrm{m}\) Number of turns per meter $n=290\ \mathrm{m}$ Current $I=3.00 \ \mathrm{A}$ Stored energy $U=2.80 \ \mathrm{J}$

Apply the formula to find stored energy in the solenoid

The stored energy in the solenoid is given by the formula: $$ U = \frac{1}{2} L I^2 $$ Where \(U\) is the stored energy, \(L\) is the inductance of the solenoid, and \(I\) is the current.

Use the inductance formula for a solenoid

The inductance of a solenoid with n turns per meter, with a length l and cross-sectional area A is given by the formula: $$ L = \frac{\mu_0 n^2 A l}{l} $$ Where \(\mu_0 = 4\pi \times 10^{-7} \ \mathrm{Tm/A}\) is the permeability of free space.

Combine the formulas for stored energy and inductance

We can substitute the formula for the inductance \(L\) from step 3 into the formula for stored energy from step 2, which gives: $$ U = \frac{1}{2} (\frac{\mu_0 n^2 A l}{l}) I^2 $$ Now, we need to solve for the cross-sectional area \(A\).

Solve the equation for the cross-sectional area A

First, we can cancel out \(l\) from the numerator and denominator in the formula: $$ U = \frac{1}{2} (\mu_0 n^2 A) I^2 $$ Now, rearrange the formula to solve for \(A\): $$ A = \frac{2U}{\mu_0 n^2 I^2} $$

Substitute given values and find the cross-sectional area

Now substitute the given values into the formula and solve for the cross-sectional area \(A\): $$ A = \frac{2(2.80 \ \mathrm{J})}{(4\pi \times 10^{-7} \ \mathrm{Tm/A})(290 \ \mathrm{m^{-1}})^2 (3.00 \ \mathrm{A})^2} $$ Calculate the value of \(A\): $$ A \approx 4.70 \times 10^{-4} \ \mathrm{m^2} $$ So, the cross-sectional area of the solenoid is approximately \(4.70 \times 10^{-4} \ \mathrm{m^2}\).