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Chapter 29: Problem 67

A long solenoid with length \(3.0 \mathrm{~m}\) and \(n=290\) turns \(/ \mathrm{m}\) carries a current of \(3.0 \mathrm{~A} .\) It stores \(2.8 \mathrm{~J}\) of energy. What is the cross-sectional area of the solenoid?

Short Answer

Expert verified
Answer: The cross-sectional area of the solenoid is approximately \(1.02 \times 10^{-3} \mathrm{~m^2}\).

Step by step solution

01

Calculate magnetic field inside solenoid

First, we need to find the magnetic field inside the solenoid (B). Use the formula for magnetic field inside a long solenoid: $$ B = \mu_0 n I $$ Plug in the values for \(\mu_0 = 4\pi × 10^{-7} \mathrm{~T m/A}\), n = 290 turns/m, and I = 3.0 A: $$ B = (4\pi × 10^{-7} \mathrm{~T m/A}) \cdot (290 \mathrm{~turns/m}) \cdot (3.0 \mathrm{~A}) $$ Calculate the magnetic field B: $$ B \approx 3.63 \times 10^{-3} \mathrm{~T} $$
02

Find volume of the solenoid

Use the formula for stored energy inside a long solenoid: $$ W = \frac{1}{2} \cdot \frac{B^2}{\mu_0} \cdot V $$ We know W = 2.8 J, B = 3.63 × 10^{-3} T, and \(\mu_0 = 4\pi × 10^{-7} \mathrm{~T m/A}\). Plug in the values and solve for volume V: $$ 2.8 \mathrm{~J} = \frac{1}{2} \cdot \frac{(3.63 \times 10^{-3} \mathrm{~T})^2}{4\pi × 10^{-7} \mathrm{~T m/A}} \cdot V $$ Calculate the volume V: $$ V \approx 3.07 \times 10^{-3} \mathrm{~m^3} $$
03

Find the cross-sectional area of the solenoid

We have the volume V and the length L of the solenoid. Use the relationship between volume, length, and cross-sectional area: $$ V = A \cdot L $$ We know V = 3.07 × 10^{-3} m³ and L = 3.0 m. Plug in the values and solve for cross-sectional area (A): $$ 3.07 \times 10^{-3} \mathrm{~m^3} = A \cdot (3.0 \mathrm{~m}) $$ Calculate the cross-sectional area A: $$ A \approx 1.02 \times 10^{-3} \mathrm{~m^2} $$ The cross-sectional area of the solenoid is approximately \(1.02 \times 10^{-3} \mathrm{~m^2}\).

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Most popular questions from this chapter

A \(100 .-V\) battery is connected in series with a \(500 .-\Omega\) resistor. According to Faraday's Law of Induction, current can never change instantaneously, so there is always some "stray" inductance. Suppose the stray inductance is \(0.200 \mu \mathrm{H}\). How long will it take the current to build up to within \(0.500 \%\) of its final value of \(0.200 \mathrm{~A}\) after the resistor is con- nected to the battery?

An emf of \(20.0 \mathrm{~V}\) is applied to a coil with an inductance of \(40.0 \mathrm{mH}\) and a resistance of \(0.500 \Omega\). a) Determine the energy stored in the magnetic field when the current reaches one fourth of its maximum value. b) How long does it take for the current to reach this value?

A long solenoid with cross-sectional area \(A_{1}\) surrounds another long solenoid with cross-sectional area \(A_{2}

A steel cylinder with radius \(2.5 \mathrm{~cm}\) and length \(10.0 \mathrm{~cm}\) rolls without slipping down a ramp that is inclined at \(15^{\circ}\) above the horizontal and has a length (along the ramp) of \(3.0 \mathrm{~m} .\) What is the induced potential difference between the ends of the cylinder at the bottom of the ramp, if the surface of the ramp points along magnetic north?

A metal loop has an area of \(0.100 \mathrm{~m}^{2}\) and is placed flat on the ground. There is a uniform magnetic field pointing due west, as shown in the figure. This magnetic field initially has a magnitude of \(0.123 \mathrm{~T}\), which decreases steadily to \(0.075 \mathrm{~T}\) during a period of \(0.579 \mathrm{~s}\). Find the potential difference induced in the loop during this time.
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