Consider a $10 c m$ long piece of a wire which carries a current of $10 A$ . Find the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece.

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Solution

$I = 10 A, a = 10 c m = 0.1 m$
$r = O P = \frac{\sqrt{3}}{2} \times 0.1 m$
$B = \frac{μ_{0} I}{4 π r} (sin ϕ_{1} + sin ϕ_{2})$
$= \frac{10^{- 7} \times 10 \times 1}{\frac{\sqrt{3}}{2} \times 0.1} = \frac{2 \times 10^{- 5}}{1.732} = 1.154 \times 10^{- 5} T = 11.54 μ T$

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Consider a 10 cm long piece of a wire which carries a current of 10 A. Find the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece.

I=10A,a=10cm=0.1mr=OP=√32×0.1mB=μ0I4πr(sinϕ1+sinϕ2)=10−7×10×1√32×0.1=2×10−51.732=1.154×10−5T=11.54μT

Consider a $10 c m$ long piece of a wire which carries a current of $10 A$ . Find the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece.

$I = 10 A, a = 10 c m = 0.1 m$
$r = O P = \frac{\sqrt{3}}{2} \times 0.1 m$
$B = \frac{μ_{0} I}{4 π r} (sin ϕ_{1} + sin ϕ_{2})$
$= \frac{10^{- 7} \times 10 \times 1}{\frac{\sqrt{3}}{2} \times 0.1} = \frac{2 \times 10^{- 5}}{1.732} = 1.154 \times 10^{- 5} T = 11.54 μ T$