Question

# Magnetic field at P due to given structure is

A
(μ4π)Iθ2R
B
μ4π2Iθ5R
C
(μ4π)5Iθ6R
D
(μ4π)2Iθ6R

Solution

## The correct option is C $$\left( \frac { { \mu }_{ \circ } }{ 4\pi } \right) \frac { 5I\theta }{ 6R }$$Given: Current=iSolution:Magnetic field due to the structure at P is given by:$$\vec{B_{p}}=\vec{B_{1}}+\vec{B_{2}}+\vec{B_{3}}+\vec{B_{4}}+\vec{B_{5}}$$But, $$\vec{B_{2}},\vec{B_{4}}=0$$( angle subtend by straight current carrying conductor is zero)$$\vec{B_{p}}=\vec{B_{1}}+\vec{B_{3}}+\vec{B_{5}}$$$$\vec{B_{1}}$$ directed downwards according to right hand thumb rule $$\vec{B_{3}}$$directed upwards according to right hand thumb rule $$\vec{B_{5}}$$directed downwards according to right hand thumb rule So,$$\vec{B_{p}}=-\vec{B_{1}}+\vec{B_{3}}-\vec{B_{5}}$$B due to a circular current carrying conductor at it's centre is given by:$$B=\frac{\mu _{0}i}{4\pi r}.\theta$$$$So,\vec{B_{p}}=-\frac{\mu _{0}i}{4\pi R}.\theta+\frac{\mu _{0}i}{4\pi 2R}.\theta -\frac{\mu _{0}i}{4\pi 3R}.\theta$$$$|\vec{B_{p}}|=|-\frac{\mu _{0}i}{4\pi R}.\theta[1-\frac{1}{2}+\frac{1}{3}]|$$$$|\vec{B_{p}}|=\frac{\mu _{0}i}{4\pi R}.\frac{5}{6}$$$$|\vec{B_{p}}|=(\frac{\mu _{0}i}{4\pi }).\frac{5i\theta }{6R}$$Hence the correct answer is CPhysicsNCERTStandard XII

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