A) \[\text{5}.0\text{x1}{{0}^{-\text{5}}}\text{ N}{{\text{A}}^{-\text{1}}}\text{ }{{\text{m}}^{-1}}\text{, upwards}\]
B) \[\text{3}.\text{4}\times \text{1}{{0}^{-\text{5}}}\text{ N}{{\text{A}}^{-\text{1}}}\text{ }{{\text{m}}^{-\text{1}}},\text{ upwards}\]
C) \[\text{1}.\text{6}\times \text{1}{{0}^{-\text{5}}}\text{ N}{{\text{A}}^{-\text{1}}}\text{ }{{\text{m}}^{-\text{1}}},\text{ downwards}\]
D) \[\text{1}.\text{6}\times \text{1}{{0}^{-\text{5}}}\text{ N}{{\text{A}}^{\text{-1}}}\text{ }{{\text{m}}^{-\text{1}}},\text{ upwards}\]
Correct Answer: B
Solution :
Magnetic field B at the centre of a coil carrying a current, I is \[{{B}_{coil}}=\frac{{{\mu }_{0}}I}{2\,r}\] (upwards) B due to wire, \[{{B}_{wire}}=\frac{{{\mu }_{0}}I}{2\pi r}\] (downwards) Magnetic field at centre C, \[{{B}_{C}}={{B}_{coil}}+{{B}_{wire}}\] \[=\frac{{{\mu }_{0}}I}{2\,r}\] (upward) \[+\frac{{{\mu }_{0}}I}{2\,\pi r}\] (downwards) \[=\frac{{{\mu }_{0}}I}{2\,r}-\frac{{{\mu }_{0}}I}{2\pi r}=\frac{{{\mu }_{0}}I}{2r}\left[ 1-\frac{1}{\pi } \right]\] upwards Given \[I=8\,A,\,r=10\times {{10}^{-2}}\,m\] \[=\frac{4\pi \times {{10}^{-7}}\times 8}{2\times 10\times {{10}^{-2}}}\left[ 1-\frac{1}{3.14} \right]\] upwards \[=3.424\times {{10}^{-5}}\]upwards
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