Question

A conductor carrying a current of $3A$ is bent into the form of an arc of a circle of radius $10cm.$ It subtends an angle of ${120}^o$ , at the centre of the circle. What is the magnetic induction at the centre of curvature of the arc?

• A. $6.2854\times {10}^{-6}wb/m^2$
• B. $4.284\times {10}^{-6}wb/m^2$
• C. $3.14\times {10}^{-6}wb/m^2$
• D. $1.57\times {10}^{-6}wb/m^2$

Answer 1

The correct option is A $6.2854\times {10}^{-6}wb/m^2$

The magnetic field at the centre of the of the circular arc making an angle$\theta$is given as

$B=\frac{{\mu }_{0}i\theta }{4\pi R}$

Substituting the values of the given parameter in the above equation we get

Taking angle in radians

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

$B=6.2854\times {10}^{-6}\frac{wb}{m^2}$