How many turns should a $10 - \mathrm { cm }$ long ideal solenoid have if it is to generate a $1.5 - \mathrm { mT }$ magnetic field when $1.0 \mathrm {~A}$ of current runs through it? $\left( \mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } \right)$
A) 12
B) 15
C) 1200
D) 120
E) 3200

Question

Quizplus · Accepted Answer

The magnetic field $B$ generated by an ideal solenoid is given by the formula $B = \mu_0 \frac{N}{L} I$, where $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A}$, $N$ is the number of turns, $L$ is the length of the solenoid in meters, and $I$ is the current in amperes. Rearranging the formula to solve for $N$, we get $N = \frac{B \cdot L}{\mu_0 \cdot I}$. Substituting the given values, $B = 1.5 \times 10^{-3} \, \mathrm{T}$, $L = 0.1 \, \mathrm{m}$, and $I = 1.0 \, \mathrm{A}$, we find $N = \frac{1.5 \times 10^{-3} \cdot 0.1}{4\pi \times 10^{-7}} \approx 120$.

How Many Turns Should A 10−cm10 - \mathrm { cm }10−cm

How Many Turns Should A $10 - \mathrm { cm }$