A conductor is carrying a current i . The magnetic field intensity at the point textO which is the common centre for three arcs is

Question

A conductor is carrying a current i . The magnetic field intensity at the point textO which is the common centre for three arcs is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-xoahcbxakuf8qqru.jpg" /> (A) (5 Âµ0 iθ /24 π R) (B) (Âµ0 iθ /24 π R) (C) (11 Âµ0 iθ /24 π R) (D) zero

Tardigrade Admin · Accepted Answer

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=(Âµ0 I/4 π r)θ Hence, net B=(Âµ0 i/4 π ) [(1/R) - (1/2 R) + (1/3 R)]θ =(5 Âµ0 iθ /24 π R)

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

$\frac{5 µ _{0} i θ}{24 π R}$

$\frac{µ _{0} i θ}{24 π R}$

$\frac{11 µ _{0} i θ}{24 π R}$

zero

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 $B = \frac{µ _{0} I}{4 π r} θ$
Hence, net $B = \frac{µ _{0} i}{4 π} [\frac{1}{R} - \frac{1}{2 R} + \frac{1}{3 R}] θ = \frac{5 µ _{0} i θ}{24 π R}$

Q. A conductor is carrying a current i . The magnetic field intensity at the point O which is the common centre for three arcs is

24πR5µ0​iθ​

24πRµ0​iθ​

24πR11µ0​iθ​