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QJEEM22Phb17b: JEE Main 2022 Physics - Class 12 JEEM22Phb17b Solution: A wire X of length carrying a current of is placed parallel to…

A wire X of length 50 cm50 \mathrm{~cm} carrying a current of 2 A2 \mathrm{~A} is placed parallel to a long wire Y\mathrm{Y} of length 5 m5 \mathrm{~m} . The wire Y\mathrm{Y} carries a current of 3 A3 \mathrm{~A} . The distance between two wires is 5 cm5 \mathrm{~cm} and currents flow in the same direction. The force acting on the wire Y\mathrm{Y} is

JEE Main 2022 (Online) 29th July Evening Shift Physics - Magnetic Effect of Current Question 74 English

Multiple Choice Options

  • A 1.2×10−5 N1.2 \times 10^{-5} \mathrm{~N} directed towards wire X\mathrm{X} .
  • B 1.2×10−4 N1.2 \times 10^{-4} \mathrm{~N} directed away from wire X\mathrm{X} .
  • C 1.2×10−4 N1.2 \times 10^{-4} \mathrm{~N} directed towards wire X\mathrm{X} .
  • D 2.4×10−5 N2.4 \times 10^{-5} \mathrm{~N} directed towards wire X\mathrm{X} .

Step-by-Step Verified Solution

JEE Main 2022 (Online) 29th July Evening Shift Physics - Magnetic Effect of Current Question 74 English Explanation

Length of wire Y is very high compared to length of wire X. So we can assume length of wire Y is infinite compare to wire X.

Magnetic field due to wire y from 5 cm from wire towards wire X is,

B=μ0I12πd=μ0×32π×0.05B = {{{\mu _0}{I_1}} \over {2\pi d}} = {{{\mu _0} \times 3} \over {2\pi \times 0.05}}

And direction of B is outward.

Now force on wire X due to wire Y,

FYX=I2B l{F_{YX}} = {I_2}B\,l

=2×μ0×32π×0.05×5= 2 \times {{{\mu _0} \times 3} \over {2\pi \times 0.05}} \times 5

=1.2×10−5 N= 1.2 \times {10^{ - 5}}\,N

Current flowing in both the wire in same direction so both wire will attract each other with equal force.

∴\therefore FYX=FXY{F_{YX}} = {F_{XY}}

∴\therefore FXY=1.2×10−5 N{F_{XY}} = 1.2 \times {10^{ - 5}}\,N , attractive force towards wire X.

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