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Question

A solenoid 95.6 cm95.6 \mathrm{~cm} long has a radius of 1.90 cm1.90 \mathrm{~cm}, a winding of 1230 turns, and carries a current of 3.58 A3.58 \mathrm{~A}. Calculate the strength of the magnetic field inside the solenoid.

Solution

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Answered 2 years ago
Answered 2 years ago
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The magnetic field inside a solenoid carrying current II and coil density nn is given by

B=μ0nI.B=\mu_0nI.

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

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

This will bring the formula to

B=μ0NIL.B=\frac{\mu_0NI}{L}.

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

B=4π10712303.580.956=5.79103 T.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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