Question

What is the magnetic field inside a coil with the following conditions: 228 number of turns, 0.952 A of current and a length of 2.29 cms.

Answer 1

Answer:

The magnetic field inside a coil is 0.0119 T.

Explanation:

Given that,

Number of turns = 228

Current = 0.952 A

Length = 2.29 cm

We need to calculate the magnetic field

Using formula of magnetic field

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

Where, B = magnetic field

l = length

I = current

N = number of turns

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times228\times0.952}{2.29\times10^{-2}}$

$B=0.0119\ T$

Hence, The magnetic field inside a coil is 0.0119 T.