Question

A piece of wire carrying a current of 6.00 A is bent in the form of a circular are of radius 10.0 cm, and it subtends an angle of 120° at the centre. Find the magnetic field B due to this piece of wire at the centre.

Answer 1

A piece of wire carrying a current of 6.00 A is bent in the form of a circular are of radius 10.0 cm, and it subtends an angle of 120° at the centre. The magnetic field B due to this piece of wire at the centre is $1.256 \times 10^{-5} T$

Explanation:

Step 1:

Given data in the question  

Current Magnitude, I = 6 A

Half circular wire radius, r = 10 cm

Angle collapsed in the middle, θ = 120° = $2 \pi / 3$

Step 2:

Magnetic field required at the center of the curvature

$B=\frac{\mu_{0} I}{2 R} \frac{\theta}{2 \pi}$

On substituting the values  ,we get

$=\frac{4 \pi \times 10^{-7} \times 6}{2 \times 10 \times 10^{-2}} \times \frac{2 \pi}{3 \times 2 \pi}$

$=\frac{24 \pi \times 10^{-7}}{2 \times 10 \times 10^{-2}} \times \frac{1}{3}$

$=\frac{24 \pi \times 10^{-7}}{6 \times 10 \times 10^{-2}}$

$=\frac{24 \pi \times 10^{-7}}{60 \times 10^{-2}}$

$=\frac{24 \pi \times 10^{-5}}{60}$

$=0.4 \pi \times 10^{-5}$

$=0.4 \times 3.14 \times 10^{-5}$

$=1.256 \times 10^{-5} T$

Answer 2

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12.56× 10^ -6 T