Question

A conductor is carrying a current $i$ . The magnetic field intensity at the point $\text{O}$ which is the common centre for three arcs is

• A. $\frac{5 µ_{0} i\theta }{24 \pi R}$
• B. $\frac{µ_{0} i\theta }{24 \pi R}$
• C. $\frac{11 µ_{0} i\theta }{24 \pi R}$
• D. zero

Answer 1

Answer: (A)

Find $B$ due to each arc at $O$ and add them using vector addition.
Since magnetic field at the centre of an arc is equal to $B=\frac{µ_{0} I}{4 \pi r}\theta $
Hence, net $B=\frac{µ_{0} i}{4 \pi } \, \left[\frac{1}{R} - \frac{1}{2 R} + \frac{1}{3 R}\right]\theta =\frac{5 µ_{0} i\theta }{24 \pi R}$