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Moving Charges and Magnetism question

2015 · Q122
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Moving Charges and Magnetism question

2015 · Q122

NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1
A wire carrying current III has the shape shown in adjoining figure.

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 69 English
Linear parts of the wire are very long and parallel to X-axis while semicircular protion of radius R is lying in Y-Z plane. Magtnetic field at pont OOO is
  1. A
    B→=−μ0I4πR(πi^+2k^)\overrightarrow B = - {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right)B=−4πRμ0​I​(πi+2k)
  2. B
    B→=μ0I4πR(πi^−2k^)\overrightarrow B = {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i - 2\widehat k} \right)B=4πRμ0​I​(πi−2k)
  3. C
    B→=μ0I4πR(πi^+2k^)\overrightarrow B = {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right)B=4πRμ0​I​(πi+2k)
  4. D
    B→=−μ0I4πR(πi^−2k^)\overrightarrow B = - {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i - 2\widehat k} \right)B=−4πRμ0​I​(πi−2k)
View written solutionFree

Correct answer: A

Magnetic field due to segment ‘1’

B1→=μ0I4πR[sin⁡90∘+sin⁡0∘](−k^)\overrightarrow {{B_1}} = {{{\mu _0}I} \over {4\pi R}}\left[ {\sin 90^\circ + \sin 0^\circ } \right]\left( { - \widehat k} \right)B1​​=4πRμ0​I​[sin90∘+sin0∘](−k)

=−μ0I4πR(k^)=B3→= {{ - {\mu _0}I} \over {4\pi R}}\left( {\widehat k} \right) = \overrightarrow {{B_3}}=4πR−μ0​I​(k)=B3​​

Magnetic field due to segment 2

B2→=μ0I4R(−i^)=−μ0I4πR(−πi^)\overrightarrow {{B_2}} = {{{\mu _0}I} \over {4R}}\left( { - \widehat i} \right) = {{ - {\mu _0}I} \over {4\pi R}}\left( { - \pi \widehat i} \right)B2​​=4Rμ0​I​(−i)=4πR−μ0​I​(−πi)

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 69 English Explanation

∴B→\therefore \overrightarrow B∴B at centre

Bc→=B1→+B2→+B3→\overrightarrow {{B_c}} = \overrightarrow {{B_1}} + \overrightarrow {{B_2}} + \overrightarrow {{B_3}} Bc​​=B1​​+B2​​+B3​​

=−μ0I4πR(πi^+2k^) = {{ - {\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right)=4πR−μ0​I​(πi+2k)

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