Find the magnetic field at point $O$ in figure

- \frac{μ_{0} I}{4 π a} (^i +^k)

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- \frac{μ_{0} I}{4 π a} (^j +^k)

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\frac{{- μ}_{0} I}{2 π a} (^j +^k)

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\frac{{- μ}_{0} I}{2 \sqrt{2} π a} (^j +^k)

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Solution

The correct option is C $\frac{{- μ}_{0} I}{2 π a} (^j +^k)$
Due to wire in the x-z plane,

Magnetic field at O= $\frac{- μ_{0} I}{4 π (a sin 45^{\circ})} (sin 45^{\circ} + sin 45^{\circ})^j = \frac{- μ_{0} I}{2 π a}^j$

Due to wire in x-y plane,

Magnetic field at O $\frac{- μ_{0} I}{4 π (a sin 45^{\circ})} (sin 45^{\circ} + sin 45^{\circ})^k = \frac{- μ_{0} I}{2 π a}^k$

Hence total magnetic field at O $= \frac{- μ_{0} I}{2 π a} (^j +^k)$

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Find the magnetic field at point O in figure

The correct option is C −μ0I2πa(^j+^k)Due to wire in the x-z plane,Magnetic field at O=−μ0I4π(asin45∘)(sin45∘+sin45∘)^j=−μ0I2πa^jDue to wire in x-y plane,Magnetic field at O−μ0I4π(asin45∘)(sin45∘+sin45∘)^k=−μ0I2πa^kHence total magnetic field at O=−μ0I2πa(^j+^k)

Find the magnetic field at point $O$ in figure