Q. 154.0( 2 Votes )

# Consider a 10 cm long piece of a wire which carries a current of 10A. 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.

Answer :

**Given:**Length of the wire piece: x = 10 cm = 0.1 m

Current in the wire piece: i= 10 A

**Formula used:**

By Biot-Savart Law:

Here, dB is the magnitude of magnetic field element, μ

_{0}is the permeability of free space and μ

_{0}= 4π × 10

^{-7}T mA

^{-1},dl is the length element, r is the distance between the current carrying wire and the required point.

We know that, magnetic field at a point on the perpendicular bisector is

Here, x is the length of the wire.

d is the distance between point C and the midpoint O.

From Pythagoras theorem,

(BC)

^{2}= (OB)

^{2}+(OC)

^{2}

∴ (OC)

^{2}= (BC)

^{2}- (OB)

^{2}

∴ (OC)

^{2}= (0.1)

^{2}– (0.05)

^{2}

∴ OC = (7.5× 10

^{-3})

^{1/2}

∴ OC = 0.086 m = d

Substituting for B we get,

∴ B = 2.32× 10

^{-5}× 0.502

∴ B = 1.16 × 10

^{-5}T

Hence the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece is 1.16 × 10

^{-5}T.

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