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QJEEM22Ph7d43: JEE Main 2022 Physics - Class 12 JEEM22Ph7d43 Solution: Two long parallel conductors S 1 and S 2 are separated by a…

Two long parallel conductors S1 and S2 are separated by a distance 10 cm and carrying currents of 4A and 2A respectively. The conductors are placed along x-axis in X-Y plane. There is a point P located between the conductors (as shown in figure).

A charge particle of 3 π\pi coulomb is passing through the point P with velocity v→=(2i^+3j^)\overrightarrow v = (2\widehat i + 3\widehat j) m/s; where i^\widehat i and j^\widehat j represents unit vector along x & y axis respectively.

The force acting on the charge particle is 4π×10−5(−xi^+2j^)4\pi \times {10^{ - 5}}( - x\widehat i + 2\widehat j) N. The value of x is :

JEE Main 2022 (Online) 27th June Evening Shift Physics - Magnetic Effect of Current Question 95 English

Multiple Choice Options

  • A 2
  • B 1
  • C 3
  • D −- 3

Step-by-Step Verified Solution

Field at P is =(μ0×i12πr1−μ0i22πr2)(−k^)= \left( {{{{\mu _0} \times {i_1}} \over {2\pi {r_1}}} - {{{\mu _0}{i_2}} \over {2\pi {r_2}}}} \right)\left( { - \widehat k} \right)

=−(μ042π×0.04−μ0×22π×0.06)k^=−μ0×2006πk^= - \left( {{{{\mu _0}4} \over {2\pi \times 0.04}} - {{{\mu _0} \times 2} \over {2\pi \times 0.06}}} \right)\widehat k = - {{{\mu _0} \times 200} \over {6\pi }}\widehat k

So, force F→=qv→×B→\overrightarrow F = q\overrightarrow v \times \overrightarrow B

=3π(2i^+3j^)×(−(μ0×2006πk^))= 3\pi \left( {2\widehat i + 3\widehat j} \right) \times \left( { - \left( {{{{\mu _0} \times 200} \over {6\pi }}\widehat k} \right)} \right)

=3π(200μ03πj^−100μ0πi^)= 3\pi \left( {{{200{\mu _0}} \over {3\pi }}\widehat j - {{100{\mu _0}} \over \pi }\widehat i} \right)

=200μ0j^−300μ0i^= 200{\mu _0}\widehat j - 300{\mu _0}\widehat i

=4π×10−5(2j^−3i^)= 4\pi \times {10^{ - 5}}(2\widehat j - 3\widehat i)

So, x=3x = 3

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