A) Zero
B) \[2\times {{10}^{-5}}T\]
C) \[4\times {{10}^{-6}}T\]
D) \[8\times {{10}^{-7}}T\]
Correct Answer: D
Solution :
Net magnetic field at mid point O \[{{B}_{net}}={{B}_{N}}+{{B}_{S}}\] where, \[{{B}_{N}}=\]magnitude of magnetic field due to \[N-\]pole and \[{{B}_{S}}=\]magnitude of magnetic field due to S-pole But, \[{{B}_{_{N}}}={{B}_{S}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{r}^{2}}}\] \[=\frac{4\pi \times {{10}^{-7}}\times 0.01}{4\pi \times {{\left( \frac{0.1}{2} \right)}^{2}}}=4\times {{10}^{-7}}T\] \[\therefore \] \[{{B}_{net}}=8\times {{10}^{-7}}T\]You need to login to perform this action.
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