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  • If the magnetic field at '' in the given figure can be written as <img alt="K*\frac{\mu_{0} i}{4 \pi d} \tan \frac{\alp

If the magnetic field at 'P' in the given figure can be written as K*\frac{\mu_{0} i}{4 \pi d} \tan \frac{\alpha}{2}       then K is :         

Option: 1 1

Option: 2 2

Option: 3 4

Option: 4 8

Answers (1)

best_answer

Let us capture the magnetic field due to any one segment :

B=\frac{\mu _{0}i}{4\pi \left ( d\sin \alpha \right )}\left ( \sin 90^{0}-\sin \left ( 90-\alpha \right ) \right )=\frac{\mu _{0}i}{4\pi \left ( d\sin \alpha \right )}\left ( 1-\cos \alpha \right )=\frac{\mu _{0}i}{4\pi d}\tan \frac{\alpha }{2}

Resultant field will be : B_{net}= 2B= \frac{\mu _{0}i}{2\pi d}\tan \frac{\alpha }{2}\, \, \, \, ,K=2

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mansi

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