Question

A solenoid of length 0.700m having a circular cross-section of radius 5.00cm stores 6.00 μJ of energy when a 0.400-A current runs through it.
What is the winding density of the solenoid? (μ0 = 4π*10-7 T*m/A)

A) 865 turns/m

B) 327 turns/m

C) 1080 turns/m

D) 104 turns/m

E) 472 turns/m

Answer 1

Answer:

The winding density of the solenoid, n = 104 turns/m

Explanation:

Given that,

Length of the solenoid, l = 0.7 m

Radius of the circular cross section, r = 5 cm = 0.05 m

Energy stored in the solenoid, $E=6\ \mu J=6\times 10^{-6}\ J$

Current, I = 0.4 A

To find,

The  winding density of the solenoid.

Solution,

The expression for the energy stored in the solenoid is given by :

$U=\dfrac{1}{2}LI^2$

Where

L is the self inductance of the solenoid

$L=\mu_on^2lA$

n is the winding density of the solenoid

$n=\sqrt{\dfrac{2U}{\mu_oI^2l\pi r^2}}$

$n=\sqrt{\dfrac{2\times 6\times 10^{-6}}{4\pi \times 10^{-7}\times 0.7\times (0.4)^2\pi (0.05)^2}}$

n = 104 turns/m

So, the winding density of the solenoid is 104 turns/m

Answer 2

Answer

The winding density of the solenoid is 104 turns/m.

(D) is correct option.

Explanation:

Given that,

Length = 0.700 m

Radius = 5.00 cm

Energy = 6.00 μJ

Current = 0.400 A

We need to calculate the density of the solenoid

Using formula of the energy store

$E=\dfrac{B^2}{2\mu_{0}}Al$

Put the value of magnetic field

$E=\dfrac{(\mu_{0}ni)^2}{2\mu_{0}}Al$

$n=\sqrt(\dfrac{2E}{\mu_{0}i^2Al})$

Put the value into the formula

$n=\sqrt{\dfrac{2\times6.00\times10^{-6}}{4\pi\times10^{-7}\times(0.4)^2\times\pi\times(5.00\times10^{-2})^2\times0.700}}$

$n=104\ turns/m$

Hence, The winding density of the solenoid is 104 turns/m.