## Related questions with answers

Question

A solenoid $95.6 \mathrm{~cm}$ long has a radius of $1.90 \mathrm{~cm}$, a winding of 1230 turns, and carries a current of $3.58 \mathrm{~A}$. Calculate the strength of the magnetic field inside the solenoid.

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 2The magnetic field inside a solenoid carrying current $I$ and coil density $n$ is given by

$B=\mu_0nI.$

Let's remember that the coil density is simply the number of coils per unit length:

$n:=\frac{N}{L}.$

This will bring the formula to

$B=\frac{\mu_0NI}{L}.$

The numerical answer can be now found by substituting. We will have

$B=\frac{4\pi \cdot 10^{-7}\cdot 1230\cdot 3.58}{0.956}=\boxed{5.79\cdot 10^{-3}~\mathrm{T}}.$

As we see, the radius was a redundant parameter in this calculation.

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