A) 300 p'
B) 500 p'
C) 400 p'
D) 100 p'
Correct Answer: B
Solution :
\[=\frac{4}{3}\times {{10}^{-5}}\] \[\therefore \] \[B=1.33\times {{10}^{-5}}T\] \[\lambda \propto \frac{1}{p}\] \[{{V}_{g}}=\frac{1}{2}\times 100mV\] \[G=\frac{{{V}_{g}}}{{{I}_{g}}}=\frac{0.05}{0.01}\] \[=5\Omega \] \[{{\text{B}}_{\text{R}}}\text{=}\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\text{.}\frac{\text{2Mcos }\!\!\theta\!\!\text{ }}{{{\text{R}}^{\text{3}}}}\] \[{{B}_{0}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{M\sin \theta }{{{R}^{3}}}\] \[\tan \phi =\frac{{{B}_{V}}}{{{B}_{H}}}=-\frac{{{B}_{B}}}{{{B}_{O}}}\] \[\frac{{{B}_{R}}}{{{B}_{O}}}=\frac{2\cos \theta }{\sin \theta }=2\cot \theta \] p = 500 p?
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