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QJEEM19Ph4c80: JEE Main 2019 Physics - Class 12 JEEM19Ph4c80 Solution: An infinitely long current carrying wire and a small current…

An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is a and distance of its centre from the wire is d (d > > a). If the loop a applies a force F on the wire then :

JEE Main 2019 (Online) 9th January Morning Slot Physics - Magnetic Effect of Current Question 167 English

Multiple Choice Options

  • A F=0F = 0
  • B F∝(ad)F \propto \left( {{a \over d}} \right)
  • C F∝(a2d3)F \propto \left( {{{{a^2}} \over {{d^3}}}} \right)
  • D F∝(ad)2F \propto {\left( {{a \over d}} \right)^2}

Step-by-Step Verified Solution

Magnetic field due to current carrying loop at distance d, is,

B = μ04π×M→×d→d3{{{\mu _0}} \over {4\pi }} \times {{\overrightarrow M \times \overrightarrow d } \over {{d^3}}}

= μ04π×i×πa2×d×sin⁡90od3{{{\mu _0}} \over {4\pi }} \times {{i \times \pi {a^2} \times d \times \sin {{90}^o}} \over {{d^3}}}    [as  M = iA]

= μ04π×iπa2d2{{{\mu _0}} \over {4\pi }} \times {{i\pi {a^2}} \over {{d^2}}}

∴\therefore    B ∝\propto a2d2{{{a^2}} \over {{d^2}}}

We also know,

F = Bil

∴\therefore    F ∝\propto B

So, F ∝\propto a2d2{{{a^2}} \over {{d^2}}}

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