Question

A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10-2 m will have what strength magnetic field at its center? (magnetic permeability in empty space µ0 = 4p · 10-7 T m/A)

Answer 1

Answer:

Therefore,

Strength magnetic field at its center, B

$B = 2.51\times 10^{-2}\ T$

Explanation:

Given:

Turn = N = 500

length of solenoid = l = 0.10 m

Current, I = 4.0 A

Radius, r = 0.01 m

To Find:

Strength magnetic field at its center, B = ?

Solution:

If N is the number of turns in the length, the total current through the rectangle is NI. Therefore, Ampere’s law applied to this path gives

$\int\ {B} \, ds = Bl=\mu_{0}NI$

Where,

B = Strength of magnetic field

l =  Length of solenoid

N = Number of turns

I = Current

$\mu_{0}=Permeability\ in\ free\ space = 4\pi\times 10^{-7}\ Tm/A$

Therefore,

$B =\dfrac{\mu_{0}NI}{l}$

Substituting the values we get

$B =\dfrac{4\times 3.14\times 10^{-7}\times 500\times 4}{0.10}=2.51\times 10^{-2}\ T$

Therefore,

Strength magnetic field at its center, B

$B = 2.51\times 10^{-2}\ T$