How much current must pass through a 400 -turn ideal solenoid that is $4.0 \mathrm {~cm}$ long to generate a 1.0-T magnetic field at the center? $\left( \mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } \right)$
A) $80 \mathrm {~A}$
B) $22 \mathrm {~A}$
C) $40 \mathrm {~A}$
D) $0.013 \mathrm {~A}$
E) $13 \mathrm {~A}$

Question

Quizplus · Accepted Answer

The magnetic field at the center of a solenoid is given by:$$B = \mu_0 n I$$where:* $B$ is the magnetic field in teslas* $\mu_0$ is the permeability of free space $(\mu_0 = 4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A})$* $n$ is the number of turns per unit length in turns per meter* $I$ is the current in amperesIn this problem, we are given:* $B = 1.0 \mathrm{~T}$* $n = 400 \mathrm{~turns} / 0.04 \mathrm{~m} = 10000 \mathrm{~turns} / \mathrm{m}$* $\mu_0 = 4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}$Solving for $I$, we get:$$I = \frac{B}{\mu_0 n} = \frac{1.0 \mathrm{~T}}{4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \times 10000 \mathrm{~turns} / \mathrm{m}} = 80 \mathrm{~A}$$Therefore, the correct answer is A: $80 \mathrm{~A}$.

How Much Current Must Pass Through a 400 -Turn Ideal 4.0 cm4.0 \mathrm {~cm}4.0 cm

How Much Current Must Pass Through a 400 -Turn Ideal $4.0 \mathrm {~cm}$