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Magnetics question

2010 · Shift 0 · Q64
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Magnetics question

2010 · Shift 0 · Q64

JEE MainPhysicsMagneticsMCQ+4 / −1
Two long parallel wires are at a distance 2d2d2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field BBB along the line XX′XX'XX′ is given by
  1. A
    AIEEE 2010 Physics - Magnetic Effect of Current Question 193 English Option 1
  2. B
    AIEEE 2010 Physics - Magnetic Effect of Current Question 193 English Option 2
  3. C
    AIEEE 2010 Physics - Magnetic Effect of Current Question 193 English Option 3
  4. D
    AIEEE 2010 Physics - Magnetic Effect of Current Question 193 English Option 4
View written solutionFree

Correct answer: A

  1. Set up the geometry

    Two long parallel wires are separated by distance 2d2d2d and both carry equal currents out of the plane.

    Let the wires be placed on the xxx-axis at: x=−dandx=+dx=-d \quad \text{and} \quad x=+dx=−dandx=+d and let the line XX′XX'XX′ be the straight line joining them and extending beyond them.

  2. Magnetic field due to one long straight wire

    For a long straight wire carrying current III, the magnetic field magnitude at perpendicular distance rrr is B=μ0I2πrB=\frac{\mu_0 I}{2\pi r}B=2πrμ0​I​

    Its direction is given by the right-hand rule.

  3. Direction of fields on the line XX′XX'XX′

    Since both currents are out of the plane, the field circles each wire anticlockwise.

    Along the line joining the wires:

    • To the left of both wires (x<−dx<-dx<−d): both fields are in the same direction, so they add.
    • Between the wires (−d<x<d-d<x<d−d<x<d): the two fields are in opposite directions, so they subtract.
    • To the right of both wires (x>dx>dx>d): both fields are again in the same direction, so they add.
  4. Expression for net field in different regions

    (i) For x>dx>dx>d

    Distances from the two wires are r1=x−d,r2=x+dr_1=x-d, \qquad r_2=x+dr1​=x−d,r2​=x+d So, B=μ0I2π(1x−d+1x+d)B=\frac{\mu_0 I}{2\pi}\left(\frac{1}{x-d}+\frac{1}{x+d}\right)B=2πμ0​I​(x−d1​+x+d1​) This is positive in magnitude and tends to infinity as x→d+x\to d^+x→d+.

    (ii) For −d<x<d-d<x<d−d<x<d

    Here the fields oppose each other. Magnitude is B=μ0I2π∣1x+d−1d−x∣B=\frac{\mu_0 I}{2\pi}\left|\frac{1}{x+d}-\frac{1}{d-x}\right|B=2πμ0​I​​x+d1​−d−x1​​ More usefully, with sign, B=μ0I2π(1x+d−1d−x)B=\frac{\mu_0 I}{2\pi}\left(\frac{1}{x+d}-\frac{1}{d-x}\right)B=2πμ0​I​(x+d1​−d−x1​) At the midpoint x=0x=0x=0, B=μ0I2π(1d−1d)=0B=\frac{\mu_0 I}{2\pi}\left(\frac{1}{d}-\frac{1}{d}\right)=0B=2πμ0​I​(d1​−d1​)=0 So the graph must cross zero exactly at the center.

    Also, as we approach either wire from inside, the field magnitude becomes very large: x→−d+⇒∣B∣→∞,x→d−⇒∣B∣→∞x\to -d^+ \Rightarrow |B|\to \infty, \qquad x\to d^- \Rightarrow |B|\to \inftyx→−d+⇒∣B∣→∞,x→d−⇒∣B∣→∞

  5. Important features of the graph

    Therefore the correct BBB vs position graph along XX′XX'XX′ must have:

    1. Vertical asymptotes at the positions of the two wires, x=−dx=-dx=−d and x=+dx=+dx=+d.
    2. A zero at the midpoint x=0x=0x=0.
    3. Symmetry about the midpoint.
    4. Outside the wires, the fields add; between the wires, they oppose.
  6. Matching with the options

    The option that shows:

    • divergence at both wires,
    • zero magnetic field at the midpoint,
    • and symmetric behavior on both sides, is Option A.
  7. Comparison with stored answer

    My derived answer is A, which matches the stored correct answer.

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