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A closely wound, circular coil with radius 2.40 cm has 800 turns. <br> (a) What must the current in the coil be if the magnetic field at the centre of the coil is0.0580 T? <br> (b) At what distance x from the centre of the coil, on the axis of the coil, is the magnetic field half its value at the centre?

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Answer Text

Answer :
A::D
Solution :
a. `B-1=(mu_0Ni)/(2R)` <br> `b. B_2 =(mu_0NiR^2)/(2(R^2+x^2)^(3/2)) given B_2=B_1/2`

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