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Chapter 26: Problem 67

A solenoid used in a plasma physics experiment is \(10 \mathrm{cm}\) in \(\mathrm{di}-\) ameter, is \(1.0 \mathrm{m}\) long, and carries a \(35-\mathrm{A}\) current to produce a \(100-\mathrm{mT}\) magnetic field. (a) How many turns are in the solenoid? (b) If the solenoid resistance is \(2.7 \Omega\), how much power

Short Answer

Expert verified
Approximately 2221 turns are needed in the solenoid to produce the given magnetic field. The power dissipated in the solenoid, calculated using Ohm's law, is approximately 3307.5W.

Step by step solution

01

Determine the Number of Turns

Ampere's law tells us that the magnetic field inside a solenoid is given by the formula \(B = \mu_0 * n * I\), where B is the magnetic field strength, \(\mu_0\) is the magnetic constant (\(4\pi * 10^{-7} \, T \cdot m/A\)), n is the number of turns per meter, and I is the current. We can solve for n: \(n = B/(\mu_0 * I)\). Plugging the given values for B (\(100 * 10^{-3}T\)) and I (35A), we find n.
02

Convert Number of Turns to Total Number of Turns

The previous step provides the number of turns per meter. To find the total number of turns in the length of the solenoid, we need to multiply this by the length of the solenoid (in meters). The total number of turns is \(N = n * length\), where the length of the solenoid is given as 1 meter.
03

Calculate the Power Dissipated

The power dissipated by a resistor can be calculated using Ohm's law as P = I^2*R, where I is the current through the resistor and R is its resistance. The exercise gives us the values for both current (35A) and resistance (2.7 Ohms).

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