4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

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Payal Gupta | Contributor-Level 10

8 months ago

4.1 Number of turns of the circular coil, n = 100

Radius of each turn, r = 8.0 cm = 0.08 m

Current flowing in the coil, I = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given as

$|\overset{?}{B}|$ = $\frac{μ_{0}}{4 π} \frac{2 π n I}{r}$ where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ Tm $A^{- 1}$
Hence, $|\overset{?}{B}|$ = $\frac{4 π * 10^{- 7}}{4 π} \frac{2 π * 100 * 0.4}{0.08}$ = 3.14 $* 10^{- 4}$ T

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