JEE Main 2022 · 27 Jun · Shift 2 · Physics Q43

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PhysicsSingle correct+4 / -1Question 43 of 90

Two long parallel conductors $S_1$ and $S_2$ 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 $3\pi$ coulomb is passing through the point P with velocity $\vec v=\left(2\hat i+3\hat j\right)$ m/s; where $\hat i$ & $\hat j$ represents unit vector along $x$ & $y$ axis respectively.

The force acting on the charge particle is $4\pi\times10^{-5}\left(-x\hat i+2\hat j\right)$ N. The value of $x$ is :

Axes X, Y, Z; conductor S1 (4 A, along +x) above conductor S2 (2 A, along +x), 10 cm apart in the X-Y plane; point P lies between them, 4 cm below S1.

-3

Answer and worked solution

Official answer: C

Fields at P (currents along $+x$ ). $S_1$ is 4 cm above P: $B_1=\dfrac{\mu_0(4)}{2\pi(0.04)}=2\times10^{-5}$ T along $-\hat k$ . $S_2$ is 6 cm below P: $B_2=\dfrac{\mu_0(2)}{2\pi(0.06)}\approx0.67\times10^{-5}$ T along $+\hat k$ . Net $\frac43\times10^{-5}$ T along $-\hat k$ .

Force. $q\vec v\times\vec B=3\pi(2,3,0)\times\left(0,0,-\tfrac43\times10^{-5}\right)=4\pi\times10^{-5}(-3,2,0)$ .

So $x=3$ .

(C)

Topics:Force on a moving charge Biot-Savart law

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JEE Main 2022 · 27 Jun · Shift 2, Physics question 43