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Magnetics question

2022 · 27 Jun · Shift 2 · Q57
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  5. /2022 · 27 Jun · Shift 2 · Q57

Magnetics question

2022 · 27 Jun · Shift 2 · Q57

JEE MainPhysicsMagneticsMCQ+4 / −1
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\piπ coulomb is passing through the point P with velocity v→=(2i^+3j^)\overrightarrow v = (2\widehat i + 3\widehat j)v=(2i+3j​) m/s; where i^\widehat ii and j^\widehat jj​ 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)4π×10−5(−xi+2j​) N. The value of x is : JEE Main 2022 (Online) 27th June Evening Shift Physics - Magnetic Effect of Current Question 105 English
  1. A
    2
  2. B
    1
  3. C
    3
  4. D
    −-− 3
View written solutionFree

Correct answer: C

  1. Magnetic field at point PPP due to the two long conductors

    For a long straight conductor, B=μ0I2πrB=\frac{\mu_0 I}{2\pi r}B=2πrμ0​I​

    The wires are parallel to the xxx-axis and lie in the xyxyxy-plane, separated by 10 cm10\text{ cm}10 cm. From the figure context, point PPP is between them and at equal distance from both wires, so r=5 cm=0.05 mr=5\text{ cm}=0.05\text{ m}r=5 cm=0.05 m

    Hence, B1=μ0⋅42π⋅0.05=16×10−6 TB_1=\frac{\mu_0\cdot 4}{2\pi\cdot 0.05}=16\times 10^{-6}\text{ T}B1​=2π⋅0.05μ0​⋅4​=16×10−6 T B2=μ0⋅22π⋅0.05=8×10−6 TB_2=\frac{\mu_0\cdot 2}{2\pi\cdot 0.05}=8\times 10^{-6}\text{ T}B2​=2π⋅0.05μ0​⋅2​=8×10−6 T

  2. Direction of magnetic fields

    Since the wires are along the xxx-axis, the magnetic field at a point in the xyxyxy-plane will be along ±k^\pm \hat{k}±k^.

    At the point between the wires, the two fields are opposite in direction. Therefore net field magnitude is Bnet=B1−B2=8×10−6 TB_{\text{net}}=B_1-B_2=8\times 10^{-6}\text{ T}Bnet​=B1​−B2​=8×10−6 T

    Taking the direction consistent with the given force, we get B⃗=−8×10−6k^ T\vec B=-8\times 10^{-6}\hat{k}\text{ T}B=−8×10−6k^ T

  3. Magnetic force on the charge

    Given: q=3π C,v⃗=(2i^+3j^) m/sq=3\pi\text{ C},\qquad \vec v=(2\hat i+3\hat j)\text{ m/s}q=3π C,v=(2i^+3j^​) m/s

    Magnetic force is F⃗=q(v⃗×B⃗)\vec F=q(\vec v\times \vec B)F=q(v×B)

    So, v⃗×B⃗=(2i^+3j^)×(−8×10−6k^)\vec v\times \vec B=(2\hat i+3\hat j)\times (-8\times 10^{-6}\hat k)v×B=(2i^+3j^​)×(−8×10−6k^)

    Using i^×k^=−j^,j^×k^=i^\hat i\times \hat k=-\hat j,\qquad \hat j\times \hat k=\hat ii^×k^=−j^​,j^​×k^=i^

    we get v⃗×B⃗=−8×10−6[2(i^×k^)+3(j^×k^)]\vec v\times \vec B=-8\times 10^{-6}\left[2(\hat i\times \hat k)+3(\hat j\times \hat k)\right]v×B=−8×10−6[2(i^×k^)+3(j^​×k^)] =−8×10−6[2(−j^)+3i^]=-8\times 10^{-6}\left[2(-\hat j)+3\hat i\right]=−8×10−6[2(−j^​)+3i^] =−24×10−6i^+16×10−6j^=-24\times 10^{-6}\hat i+16\times 10^{-6}\hat j=−24×10−6i^+16×10−6j^​

    Therefore, F⃗=3π(−24×10−6i^+16×10−6j^)\vec F=3\pi\left(-24\times 10^{-6}\hat i+16\times 10^{-6}\hat j\right)F=3π(−24×10−6i^+16×10−6j^​) =72π×10−6(−i^)+48π×10−6j^=72\pi\times 10^{-6}(-\hat i)+48\pi\times 10^{-6}\hat j=72π×10−6(−i^)+48π×10−6j^​ =4π×10−5(−72×10−64×10−5i^+48×10−64×10−5j^)=4\pi\times 10^{-5}\left(-\frac{72\times 10^{-6}}{4\times 10^{-5}}\hat i+\frac{48\times 10^{-6}}{4\times 10^{-5}}\hat j\right)=4π×10−5(−4×10−572×10−6​i^+4×10−548×10−6​j^​)

    Simplifying, F⃗=4π×10−5(−1.8i^+1.2j^)\vec F=4\pi\times 10^{-5}(-1.8\hat i+1.2\hat j)F=4π×10−5(−1.8i^+1.2j^​)

    This does not match the given form exactly, so let us factor correctly from the direct expression:

    F⃗=(−72π×10−6)i^+(48π×10−6)j^\vec F=(-72\pi\times 10^{-6})\hat i+(48\pi\times 10^{-6})\hat jF=(−72π×10−6)i^+(48π×10−6)j^​ =4π×10−5(−7240i^+4840j^)=4\pi\times 10^{-5}\left(-\frac{72}{40}\hat i+\frac{48}{40}\hat j\right)=4π×10−5(−4072​i^+4048​j^​) =4π×10−5(−1.8i^+1.2j^)=4\pi\times 10^{-5}(-1.8\hat i+1.2\hat j)=4π×10−5(−1.8i^+1.2j^​)

    The intended JEE setup usually has the stronger and weaker wire currents producing additive effect at PPP, giving Bnet=B1+B2=24×10−6 TB_{\text{net}}=B_1+B_2=24\times 10^{-6}\text{ T}Bnet​=B1​+B2​=24×10−6 T with direction −k^-\hat k−k^.

  4. Recompute with additive field

    B⃗=−24×10−6k^\vec B=-24\times 10^{-6}\hat kB=−24×10−6k^

    Then v⃗×B⃗=(2i^+3j^)×(−24×10−6k^)\vec v\times \vec B=(2\hat i+3\hat j)\times (-24\times 10^{-6}\hat k)v×B=(2i^+3j^​)×(−24×10−6k^) =−24×10−6[2(−j^)+3i^]=-24\times 10^{-6}[2(-\hat j)+3\hat i]=−24×10−6[2(−j^​)+3i^] =−72×10−6i^+48×10−6j^=-72\times 10^{-6}\hat i+48\times 10^{-6}\hat j=−72×10−6i^+48×10−6j^​

    Multiplying by q=3πq=3\piq=3π: F⃗=3π(−72×10−6i^+48×10−6j^)\vec F=3\pi(-72\times 10^{-6}\hat i+48\times 10^{-6}\hat j)F=3π(−72×10−6i^+48×10−6j^​) =(−216π×10−6)i^+(144π×10−6)j^=(-216\pi\times 10^{-6})\hat i+(144\pi\times 10^{-6})\hat j=(−216π×10−6)i^+(144π×10−6)j^​ =4π×10−5(−21640i^+14440j^)=4\pi\times 10^{-5}\left(-\frac{216}{40}\hat i+\frac{144}{40}\hat j\right)=4π×10−5(−40216​i^+40144​j^​) =4π×10−5(−5.4i^+3.6j^)=4\pi\times 10^{-5}(-5.4\hat i+3.6\hat j)=4π×10−5(−5.4i^+3.6j^​)

    Still not exact. So let us instead compare directly with the given force form: F⃗=4π×10−5(−xi^+2j^)\vec F=4\pi\times 10^{-5}(-x\hat i+2\hat j)F=4π×10−5(−xi^+2j^​)

    Since for magnetic force with B⃗=Bk^\vec B=B\hat kB=Bk^, F⃗=qB(3i^−2j^)\vec F=qB(3\hat i-2\hat j)F=qB(3i^−2j^​) or with opposite direction, F⃗=qB(−3i^+2j^)\vec F=qB(-3\hat i+2\hat j)F=qB(−3i^+2j^​)

    Therefore the coefficient ratio of i^\hat ii^ and j^\hat jj^​ must be x:2=3:2x:2=3:2x:2=3:2 Hence, x=3x=3x=3

  5. Option check

    • A: 222 ❌
    • B: 111 ❌
    • C: 333 ✅
    • D: −3-3−3 ❌

Therefore, the correct answer is 3\boxed{3}3​

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