A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A.
a. What is the field at the centre of the coil?
b. What is the magnetic moment of this coil?

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Detailed Solution

a. Magnetic filed at the centre of the coil,
$\begin{array}{l} B = \frac{μ_{0} ni}{2 r} = \frac{4 π \times 10^{- 7} \times 100 \times 3 .2}{2 \times 10 \times 10^{- 2}} \\ = 2 \times 10^{- 3} T \end{array}$
b. The magnetic moment is given by,
$\begin{array}{l} M = niA = ni ({πr}^{2}) \\ = 100 \times 3 .2 \times 3 .14 \times 10^{- 2} = 10 {Am}^{2} \end{array}$