A) \[\frac{{{I}_{e}}R}{{{I}_{c}}\pi }\]
B) \[\frac{{{I}_{\operatorname{c}}}R}{{{I}_{\operatorname{e}}}\pi }\]
C) \[\frac{\pi {{I}_{\operatorname{c}}}}{{{I}_{\operatorname{e}}}R}\]
D) \[\frac{{{I}_{\operatorname{c}}}\pi }{{{I}_{\operatorname{e}}}R}\]
Correct Answer: A
Solution :
Magnetic field due to closed loop of radius R \[{{\vec{B}}_{1}}=\frac{{{\mu }_{\operatorname{o}}}{{\operatorname{I}}_{c}}}{2\operatorname{R}}~~\odot Out of the plane\] Magnetic field due to straight wire carrying currently \[{{\vec{B}}_{2}}=\frac{{{\mu }_{\operatorname{o}}}{{\operatorname{I}}_{c}}}{2\pi \operatorname{H}}~~\otimes \operatorname{Into} the plane\] It is given that \[{{\vec{B}}_{net}}\]at loop centre is zero \[{{\vec{B}}_{1}}-{{\vec{B}}_{2}} =0\] \[{{\vec{B}}_{1}}={{\vec{B}}_{2}}\] \[\frac{{{\mu }_{\operatorname{o}}}{{\operatorname{I}}_{c}}}{2\operatorname{R}}=\frac{{{\mu }_{\operatorname{o}}}{{\operatorname{I}}_{c}}}{2\pi \operatorname{H}}\] \[\operatorname{H}=\frac{\operatorname{R}{{\operatorname{I}}_{c}}}{\pi {{\operatorname{I}}_{c}}}\]
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