question_answer

A coil having \[N\] turns carry a current as shown in the figure. The magnetic field intensity at point \[P\] is:                                                                                     [BHU M-2003]

A)  \[\frac{{{\mu }_{0}}Ni{{R}^{2}}}{2{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}\]                   

B)  \[\frac{{{\mu }_{0}}Ni}{2\,R}\]

C)   \[\frac{{{\mu }_{0}}Ni{{R}^{2}}}{{{\left( R+x \right)}^{2}}}\]                      

D)  \[Zero\]

Correct Answer: A

Solution :

                 Key Idea: The resultant magnetic field at P is found only along the axis. Magnetic field being a vector quantity can be resolved into two components, one of magnitude \[\delta B\,\,\sin \,\,\phi \] along the axis of the coil and \[\delta B\,\,\cos \,\,\phi \] at right angles. The resultant field at P is divided along the axis and its magnitude is given by                                                 \[B=\int{dB\,\sin \,\phi }\] For current \[i\] flowing in the coil of radius\[r\].                                                 \[B=\frac{{{\mu }_{0}}}{4\,\pi }\frac{i}{{{r}^{2}}}\int{dl\,\,\sin \,\,\phi }\] From figure, \[\sin \,\phi =\frac{R}{r}\]                                 \[B=\frac{{{\mu }_{0}}}{4\,\pi }\frac{i\,R}{{{r}^{3}}}\int{dl}\] Also, \[\int{dl=2\,\pi \,R}\,and\,r={{\left( {{R}^{2}}+{{x}^{2}} \right)}^{1/2}}\] \[\therefore B=\frac{{{\mu }_{0}}}{4\,\pi }\frac{2\,\pi \,i\,{{R}^{2}}}{{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}=\frac{{{\mu }_{0}}\,i\,{{R}^{2}}}{2{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}\] Since, coil has \[N\] turns, therefore                                 \[B=\frac{{{\mu }_{0}}\,Ni\,{{R}^{2}}}{2{{\left( {{R}^{2}}+{{x}^{2}}0 \right)}^{3/2}}}\]