Question

The figure below shows a current loop having two circular arcs joined by two radial lines. The magnetic field at O is

• A. $\frac{{\mu }_{0}i\theta }{2\pi ab}(b-a)$
• B. $\frac{{\mu }_{0}i\theta }{4\pi ab}(b-a)$
• C. Zero
• D. $\frac{{\mu }_{0}i\theta }{3\pi ab}(b-a)$

Answer 1

The correct option is B $\frac{{\mu }_{0}i\theta }{4\pi ab}(b-a)$

Let the magnetic field at point O because of arc 1 & arc 2 be $B_1$& $B_{2}dB=\frac{{\mu }_{0}Idlsin\alpha }{4\pi r^2}\alpha ={90}^{\circ }andsin{90}^{\circ }=1dB=\frac{{\mu }_{0}Idl}{4\pi r}{B}_1=\frac{{\mu }_{0}idl}{4\pi a^2}dl\to arclengthdl=r\theta =a\theta B_1=\frac{{\mu }_{0}ia\theta }{4\pi a^2}=\frac{{\mu }_{0}i\theta }{4\pi a}$

$B_1=\frac{{\mu }_{0}i\theta }{4\pi a}\to \text{out of the plane}{B}_2=\frac{{\mu }_{0}idl}{4\pi b^2}dl=b\theta B_2=\frac{{\mu }_{0}ib\theta }{4\pi b^2}=\frac{{\mu }_{0}i\theta }{4\pi b}{B}_2=\frac{{\mu }_{0}i\theta }{4\pi b}\to \text{into the plane}\text{Net magnetic field at point O}B=B_1-B_2=\frac{{\mu }_{0}i\theta }{4\pi a}-\frac{{\mu }_{0}i\theta }{4\pi b}=\frac{{\mu }_{0}i\theta }{4\pi }[\frac{1}{a}-\frac{1}{b}]=\frac{{\mu }_{0}i\theta }{4\pi }\left[\frac{b-a}{ab}\right]B=\frac{{\mu }_{0}i\theta }{4\pi ab}[b-a]\to \text{out of the plane}$