A) \[\frac{{{\mu }_{0}}I}{4r}\]
B) \[\frac{{{\mu }_{0}}I}{2r}\]
C) \[\frac{{{\mu }_{0}}I}{2\pi r}(\pi +1)\]
D) \[\frac{{{\mu }_{0}}I}{2\pi r}(\pi -1)\]
Correct Answer: C
Solution :
The magnitude of the magnetic field at point \[O\] due to straight part of wire is \[{{B}_{1}}\]is perpendicular to the plane of the page, directed upwards (right hand palm rule1). The field at the centre 0 due to the current loop of radius \[r\] is \[{{B}_{2}}=\frac{{{\mu }_{0}}i}{2r}\] \[{{B}_{2}}\]is also perpendicular to the page, directed upwards (right hand screw rule). \[\therefore \]Resultant field at \[O\] is \[{{B}_{1}}+{{B}_{2}}=\frac{{{\mu }_{0}}i}{2r}\left( \frac{1}{\pi }+1 \right)\] \[=\frac{{{\mu }_{0}}i}{2\pi r}(\pi +1)\]You need to login to perform this action.
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