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QJEEM23Ph8bcb: JEE Main 2023 Physics - Class 12 JEEM23Ph8bcb Solution: A current carrying rectangular loop PQRS is made of uniform…

A current carrying rectangular loop PQRS is made of uniform wire. The length PR=QS=5 cmP R=Q S=5 \mathrm{~cm} and PQ=RS=100 cmP Q=R S=100 \mathrm{~cm} . If ammeter current reading changes from I to 2I2 I , the ratio of magnetic forces per unit length on the wire PQP Q due to wire RSR S in the two cases respectively (fPQI:fPQ2t)\left(f_{P Q}^I: f_{P Q}^{2 t}\right) is:

JEE Main 2023 (Online) 30th January Evening Shift Physics - Magnetic Effect of Current Question 67 English

Multiple Choice Options

  • A 1:04
  • B 1:03
  • C 1:02
  • D 1:05

Step-by-Step Verified Solution

Force between two current carrying wire

=μ0I1I22πd×L= {{{\mu _0}{I_1}{I_2}} \over {2\pi d}} \times L

Here, I1{I_1} & I2{I_2} are equal

F=μ0I22πd×LF = {{{\mu _0}{I^2}} \over {2\pi d}} \times L

F∝I2F \propto {I^2}

FIF2I=I24I2=14{{{F_I}} \over {{F_{2I}}}} = {{{I^2}} \over {4{I^2}}} = {1 \over 4}

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