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A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur

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samuelonum1

Answer:

The current required  winding is  [tex]2.65*10^-^2 mA[/tex]

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil  and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= [tex]4\pi*10^-^7 T.m/A[/tex]

Applying the equation  B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

[tex]I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }[/tex]

[tex]I= 2.65*10^-^2 mA[/tex]

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