Question

Magnetic field at point O due to the structure is

• A. $\frac{{\mu }_{0}I}{4\pi R}(\frac{\pi }{R}+1)\otimes$
• B. $\frac{{\mu }_{0}I}{4\pi R}(\pi +1)\odot$
• C. $\frac{{\mu }_{0}I}{2\pi R}(\frac{\pi }{R}+1)\otimes$
• D. $\frac{{\mu }_{0}I}{2\pi R}(\pi +1)\odot$

Answer 1

The correct option is A $\frac{{\mu }_{0}I}{4\pi R}(\frac{\pi }{R}+1)\otimes$

Let the magnetic field at point O because of point 1, 2 and 3 be $B_1,B_{2}and{B}_3{B}_1=0\left(PointOliesontheconductor\right)B_2=\frac{{\mu }_{0}Idl}{4\pi R^2}{B}_2=\frac{{\mu }_{0}I\left(\frac{\pi R}{2}\right)}{4\pi R^2}=\frac{{\mu }_{0}I\pi R}{8\pi R^2}{B}_2=\frac{{\mu }_{0}I}{8R}\otimes \to \text{into the plane}{B}_3=\frac{{\mu }_{0}I}{4\pi R}(\text{point O lies close to the one}\text{end of the conductor})B_3=\frac{{\mu }_{0}I}{4\pi R}\otimes \to \text{into the plane}\text{So, net magnetic field at point O}B=B_1+B_2+B_3=0+\frac{{\mu }_{0}I}{8R}+\frac{{\mu }_{0}I}{4\pi R}B=\frac{{\mu }_{0}I}{4\pi R}[\frac{\pi }{2}+1]\otimes \to \text{Into the plane}$