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A closely wound, circular coil with radius 2.80 cm has 790 turns. Part A What must the current in the coil be if the magnetic field at the center of the coil is 0.0760 T

Sagot :

okpalawalter8

Answer:

[tex]I=11.1A[/tex]

Explanation:

From the question we are told that:

Radius [tex]R=2.80[/tex]

Turns [tex]N=790[/tex]

Magnetic field B=0.0760

Generally the equation for Magnetic field at the center of the coil is mathematically given by

 [tex]B=\frac{\mu NI}{2r}[/tex]

 [tex]0.076=\frac{4\p*10^-^7*790*I}{2*0.028}[/tex]

 [tex]I=\frac{0.076*2*0.028}{4\p*10^{-7}*790}[/tex]

 [tex]I=11.1A[/tex]

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