Question
A battery is connected between two points $$A$$ and $$B$$ on the circumference of a uniform conducting ring of radius $$r$$ and resistance $$R.$$ One of the arcs $$AB$$ of the ring subtends an angle $$\theta $$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is
A.
proportional to $$2\left( {{{180}^ \circ } - \theta } \right)$$
B.
inversely proportional to $$r$$
C.
zero, only if $$\theta = {180^ \circ }$$
D.
zero for all values of $$\theta $$
Answer :
zero for all values of $$\theta $$
Solution :
Magnetic field at the centre due to current in arc $$ABC$$ is
$${B_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_1}}}{r}\theta \,\,\,\,\,\,\,\,\left( {{\text{Directed}}\,{\text{upwards}}} \right)$$
Magnetic field at the centre due to current in arc $$ADB$$ is
$${B_2} = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_2}}}{r}\left( {2\pi - \theta } \right)\,\,\,\,\,\,\left( {{\text{Directed}}\,{\text{downwards}}} \right)$$
Therefore net magnetic field at the centre
$$\eqalign{ & B = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_1}}}{r}\frac{\theta }{\pi } - \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_2}}}{r}\left( {2\pi - \theta } \right) \cr & {\text{Also}},\,{I_1} = \frac{E}{{{R_1}}} = \frac{E}{{\frac{{\rho {\ell _1}}}{A}}} = \frac{{EA}}{{\rho r\theta }} \cr & {\text{and}}\,{I_2} = \frac{E}{{{R_2}}} = \frac{E}{{\frac{{\rho {\ell _2}}}{A}}} = \frac{{EA}}{{\rho r\left( {2\pi - \theta } \right)}} \cr & \therefore B = \frac{{{\mu _0}}}{{4\pi }}\left[ {\frac{{EA}}{{\rho r\theta }} \times \frac{\theta }{r} - \frac{{EA}}{{\rho r\left( {2\pi - \theta } \right)}} \times \frac{{\left( {2\pi - \theta } \right)}}{r}} \right] = 0 \cr} $$
Magnetic field at the centre due to current in arc $$ABC$$ is
$${B_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_1}}}{r}\theta \,\,\,\,\,\,\,\,\left( {{\text{Directed}}\,{\text{upwards}}} \right)$$
Magnetic field at the centre due to current in arc $$ADB$$ is
$${B_2} = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_2}}}{r}\left( {2\pi - \theta } \right)\,\,\,\,\,\,\left( {{\text{Directed}}\,{\text{downwards}}} \right)$$
Therefore net magnetic field at the centre
$$\eqalign{ & B = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_1}}}{r}\frac{\theta }{\pi } - \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_2}}}{r}\left( {2\pi - \theta } \right) \cr & {\text{Also}},\,{I_1} = \frac{E}{{{R_1}}} = \frac{E}{{\frac{{\rho {\ell _1}}}{A}}} = \frac{{EA}}{{\rho r\theta }} \cr & {\text{and}}\,{I_2} = \frac{E}{{{R_2}}} = \frac{E}{{\frac{{\rho {\ell _2}}}{A}}} = \frac{{EA}}{{\rho r\left( {2\pi - \theta } \right)}} \cr & \therefore B = \frac{{{\mu _0}}}{{4\pi }}\left[ {\frac{{EA}}{{\rho r\theta }} \times \frac{\theta }{r} - \frac{{EA}}{{\rho r\left( {2\pi - \theta } \right)}} \times \frac{{\left( {2\pi - \theta } \right)}}{r}} \right] = 0 \cr} $$