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What current is required in the windings of a long solenoid that has 1000 turns uniformly distributed over a length of 0.4 m, to produce at the center of the solenoid a magnetic field of magnitude 10000 T.

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Answer:


I=3.18*10^(6) A

Step-by-step explanation:

The solenoid is defined as a cylindrical coil that has a wire of conductive material wound on it, so that, with the passage of electric current, an intense magnetic field is generated. When this magnetic field appears, it begins to operate like a magnet; The magnetic field is comparable to that of a straight magnet.

The magnetic field of a solenoid is calculated using the formula:

B= µo*I*
(N)/(L) Equation 1

Where:

B: magnetic field in Teslas (T)

µo: free space permeability in T*m/A

I= Intensity of the current flowing through the conductor in ampere (A)

N= number of turns

L= solenoid length in meters (m)

Data of the problem:

L=0.4m, B= 10000T, N=1000 turns, µo=
4*\pi *10^(-7)

We cleared I of the equation (1):

I=L*B/ µo*N


I=(0.4*10000)/(4\pi*10^(-7) *1000 )


I=(4)/(4\pi *10^(-7) )


I=(10^(7) )/(\pi )


I=0,318*10^(7)


I=3.18*10^(6) A

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