Question

a solenoid of length 12.1 cm carries a current of 3.19 a and has a magnetic field at the center of 1.45 × 10-3 t. how many turns does the solenoid have? 29.4 solenoids and toroids

Explanation:

Step1: Recall the formula for the magnetic field inside a solenoid

The formula for the magnetic field \(B\) inside a solenoid is \(B = \mu_0\frac{N}{L}I\), where \(\mu_0=4\pi\times 10^{-7}\ T\cdot m/A\), \(N\) is the number of turns, \(L\) is the length of the solenoid, and \(I\) is the current. We need to solve for \(N\), so we can rewrite the formula as \(N=\frac{BL}{\mu_0I}\).

Step2: Convert the length to meters

Given \(L = 12.1\ cm=0.121\ m\), \(B = 1.45\times 10^{-3}\ T\), and \(I = 3.19\ A\).

Step3: Substitute the values into the formula

Substitute \(\mu_0 = 4\pi\times 10^{-7}\ T\cdot m/A\), \(B = 1.45\times 10^{-3}\ T\), \(L = 0.121\ m\), and \(I = 3.19\ A\) into \(N=\frac{BL}{\mu_0I}\).

$$
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$$N=\frac{0.17545}{12.76\times3.1416}\times10^{4}$$

$$N=\frac{0.17545}{40.19}\times10^{4}$$

$$N = 436.5\approx437$$

Answer:

437