Question

A 40.0 mA current is carried by a uniformly wound air-core solenoid with 435 turns, a 16.5 mm diameter, and 12.5 cm length. (a) Compute the magnetic field inside the solenoid. µT (b) Compute the magnetic flux through each turn. T·m2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current? (Select all that apply.) magnetic field inside the solenoid magnetic flux through each turn inductance of the solenoid

Answer 1

Answer:

55.68 μT

$1.19058\times 10^{-8}\ Tm^2$

0.12 mH

Magnetic flux and magnetic field

Explanation:

i = Current = 40 mA

N = Number of turns = 435

D = Diameter of solenoid = 16.5 mm

l = Length of wire = 12.5 cm

A = Area of solenoid = $\pi \left(\frac{D}{2}\right)^2$

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

Magnetic field of a solenoid is given by

$B=\frac{\mu_0Ni}{l}\\\Rightarrow B=\frac{4\times 10^{-7}\times 435\times 4\times 10^{-3}}{12.5\times 10^{-2}}\\\Rightarrow B=0.00005568\ T=55.68\times 10^{-6}\ T=55.68\ \mu T$

The magnetic field inside the solenoid is 55.68 μT

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00005568\times \pi\times \left(\frac{16.5\times 10^{-3}}{2}\right)^2\\\Rightarrow \phi=1.19058\times 10^{-8}\ Tm^2$

The magnetic flux through each turn is $1.19058\times 10^{-8}\ Tm^2$

Inductance of solenoid is given by

$L=\frac{N\phi}{i}\\\Rightarrow L=\frac{435\times 1.19058\times 10^{-8}}{40\times 10^{-3}}\\\Rightarrow L=0.00012\ H=0.12\ mH$

The inductance of the solenoid is 0.12 mH

As can be seen from the formulae above magnetic flux and magnetic field depend on current.