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Question

Magnetic field at P due to given structure is 
1266425_c58fb0b95f194919bb85ea405441a6a5.png


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

The correct option is C $$\left( \frac { { \mu }_{ \circ } }{ 4\pi } \right) \frac { 5I\theta }{ 6R } $$
Given: Current=i
Solution: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 C

Physics
NCERT
Standard XII

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