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QAIPM-2015-Ph-1334: NEET 2015 Physics - Class 12 AIPM-2015-Ph-1334 Solution: A wire carrying current I has the shape shown in adjoining…

A wire carrying current I has the shape shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is B⃗=−μ0I4πR(πi^+2k^)\vec{B} = -\frac{\mu_0 I}{4\pi R}(\pi\hat{i} + 2\hat{k}) B⃗=μ0I4πR(πi^−2k^)\vec{B} = \frac{\mu_0 I}{4\pi R}(\pi\hat{i} - 2\hat{k}) B⃗=μ0I4πR(πi^+2k^)\vec{B} = \frac{\mu_0 I}{4\pi R}(\pi\hat{i} + 2\hat{k}) B⃗=−μ0I4πR(πi^−2k^)\vec{B} = -\frac{\mu_0 I}{4\pi R}(\pi\hat{i} - 2\hat{k})

Multiple Choice Options

  • A B→=−μ0I4πR(πi^+2k^)\overrightarrow B = - {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right) B = _ 4 � R _ 0 _ I _ ( � i + 2 k )
  • B B→=μ0I4πR(πi^−2k^)\overrightarrow B = {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i - 2\widehat k} \right) B = 4 � R _ 0 _ I _ ( � i _ 2 k )
  • C B→=μ0I4πR(πi^+2k^)\overrightarrow B = {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right) B = 4 � R _ 0 _ I _ ( � i + 2 k )
  • D B→=−μ0I4πR(πi^−2k^)\overrightarrow B = - {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i - 2\widehat k} \right) B = _ 4 � R _ 0 _ I _ ( � i _ 2 k )

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