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QJEEM20Phf3cc: JEE Main 2020 Physics - Class 12 JEEM20Phf3cc Solution: A very long wire ABDMNDC is shown in figure carrying current I. …

A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is : JEE Main 2020 (Online) 8th January Evening Slot Physics - Magnetic Effect of Current Question 142 English

Multiple Choice Options

  • A μ0I2πR(π+1){{{\mu _0}I} \over {2\pi R}}\left( {\pi + 1} \right)
  • B μ0I2πR(π−12){{{\mu _0}I} \over {2\pi R}}\left( {\pi - {1 \over {\sqrt 2 }}} \right)
  • C μ0I2R{{{\mu _0}I} \over {2R}}
  • D μ0I2πR(π+12){{{\mu _0}I} \over {2\pi R}}\left( {\pi + {1 \over {\sqrt 2 }}} \right)

Step-by-Step Verified Solution

JEE Main 2020 (Online) 8th January Evening Slot Physics - Magnetic Effect of Current Question 142 English Explanation
B = μ0I4πd(sin⁡90∘−sin⁡45∘)⊗{{{\mu _0}I} \over {4\pi d}}\left( {\sin 90^\circ - \sin 45^\circ } \right) \otimes

+ μ0I2R⊙{{{\mu _0}I} \over {2R}} \odot + μ0I4πR(sin⁡45∘+sin⁡90∘)⊙{{{\mu _0}I} \over {4\pi R}}\left( {\sin 45^\circ + \sin 90^\circ } \right) \odot

= μ0I2πR(sin⁡45∘+π){{{\mu _0}I} \over {2\pi R}}\left( {\sin 45^\circ + \pi } \right)

= μ0I2πR(12+π){{{\mu _0}I} \over {2\pi R}}\left( {{1 \over {\sqrt 2 }} + \pi } \right)

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