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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

Chapter 14 discussed damped harmonic oscillators, in which the damping force is velocity dependent and always opposes the motion of the oscillator. One way of producing this type of force is to use a piece of metal, such as aluminum, that moves through a nonuniform magnetic field. Explain why this technique is capable of producing a damping force.

A square conducting loop with sides of length \(L\) is rotating at a constant angular speed, \(\omega\), in a uniform magnetic field of magnitude \(B\). At time \(t=0\), the loop is oriented so that the direction normal to the loop is aligned with the magnetic field. Find an expression for the potential difference induced in the loop as a function of time.

An 8 -turn coil has square loops measuring \(0.200 \mathrm{~m}\) along a side and a resistance of \(3.00 \Omega\). It is placed in a magnetic field that makes an angle of \(40.0^{\circ}\) with the plane of each loop. The magnitude of this field varies with time according to \(B=1.50 t^{3}\), where \(t\) is measured in seconds and \(B\) in teslas. What is the induced current in the coil at \(t=2.00 \mathrm{~s} ?\)

A circuit contains a 12.0 -V battery, a switch, and a light bulb connected in series. When the light bulb has a current of 0.100 A flowing in it, it just starts to glow. This bulb draws \(2.00 \mathrm{~W}\) when the switch has been closed for a long time. The switch is opened, and an inductor is added to the circuit, in series with the bulb. If the light bulb begins to glow \(3.50 \mathrm{~ms}\) after the switch is closed again, what is the magnitude of the inductance? Ignore any time to heat the filament, and assume that you are able to observe a glow as soon as the current in the filament reaches the 0.100 - A threshold.

A motor has a single loop inside a magnetic field of magnitude \(0.87 \mathrm{~T}\). If the area of the loop is \(300 \mathrm{~cm}^{2}\), find the maximum angular speed possible for this motor when connected to a source of emf providing \(170 \mathrm{~V}\).
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