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Moving Charges and Magnetism question

2010 · Q75
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Moving Charges and Magnetism question

2010 · Q75

NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1
A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is iii. The resultant magnetic field due to the two semicircular parts at their common centre is
  1. A
    μ0i22R{{{\mu _0}i} \over {2\sqrt 2 R}}22​Rμ0​i​
  2. B
    μ0i2R{{{\mu _0}i} \over {2R}}2Rμ0​i​
  3. C
    μ0i4R{{{\mu _0}i} \over {4R}}4Rμ0​i​
  4. D
    μ0i2R{{{\mu _0}i} \over {\sqrt 2 R}}2​Rμ0​i​
View written solutionFree

Correct answer: A

Magnetic fields due to the two parts at their common centre are respectively,

By=μ0i4R{B_y} = {{{\mu _0}i} \over {4R}}By​=4Rμ0​i​ and Bz=μ0i4R{B_z} = {{{\mu _0}i} \over {4R}}Bz​=4Rμ0​i​

AIPMT 2010 Mains Physics - Moving Charges and Magnetism Question 49 English Explanation

Resultant field = By2+Bz2\sqrt {B_y^2 + B_z^2} By2​+Bz2​​

= (μ0i4R)2+(μ0i4R)2\sqrt {{{\left( {{{{\mu _0}i} \over {4R}}} \right)}^2} + {{\left( {{{{\mu _0}i} \over {4R}}} \right)}^2}} (4Rμ0​i​)2+(4Rμ0​i​)2​

=2μ0i4R=μ0i22R = \sqrt 2 {{{\mu _0}i} \over {4R}} = {{{\mu _0}i} \over {2\sqrt 2 R}}=2​4Rμ0​i​=22​Rμ0​i​

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