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Magnetism question

2014 · Shift 2 · Q52
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Magnetism question

2014 · Shift 2 · Q52

JEE AdvancedPhysicsMagnetismMCQ+3 / −1
The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counter clockwise direction if seen from above. JEE Advanced 2014 Paper 2 Offline Physics - Magnetism Question 27 English ComprehensionWhen d ≈\approx≈ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case
  1. A
    current in wire 1 and wire 2 is the direction PQ and RS, respectively, and h ≈\approx≈ a.
  2. B
    current in wire 1 and wire 2 is the direction PQ and SR, respectively, and h ≈\approx≈ a.
  3. C
    current in wire 1 and wire 2 is the direction PQ and SR, respectively, and h ≈\approx≈ 1.2a.
  4. D
    current in wire 1 and wire 2 is the direction PQ and RS, respectively, and h ≈\approx≈ 1.2a.
View written solutionFree

Correct answer: C

  1. Magnetic field due to the circular loop on its axis

The loop lies in the plane of the paper, and its current is counterclockwise when seen from above. Hence by the right-hand rule, the magnetic field on the axis at a height hhh above the center is upward.

Its magnitude is

Bloop=μ0Ia22(a2+h2)3/2.B_{\text{loop}}=\frac{\mu_0 I a^2}{2(a^2+h^2)^{3/2}}.Bloop​=2(a2+h2)3/2μ0​Ia2​.
  1. Field due to the two long straight wires at a point on the axis

The two wires are parallel to each other and lie in the plane of the paper, each at distance ddd from the center. Since the observation point is at height hhh above the center, the perpendicular distance from either wire to that point is

r=d2+h2.r=\sqrt{d^2+h^2}.r=d2+h2​.

For a long straight wire,

B=μ0I2πr.B=\frac{\mu_0 I}{2\pi r}.B=2πrμ0​I​.

But we need only the vertical component (along the axis of the loop). From geometry,

Bz=B⋅dr=μ0I2πr⋅dr=μ0Id2π(d2+h2).B_z = B\cdot \frac{d}{r} = \frac{\mu_0 I}{2\pi r}\cdot \frac{d}{r} = \frac{\mu_0 I d}{2\pi(d^2+h^2)}.Bz​=B⋅rd​=2πrμ0​I​⋅rd​=2π(d2+h2)μ0​Id​.

So each wire contributes vertical component of magnitude

μ0Id2π(d2+h2).\frac{\mu_0 I d}{2\pi(d^2+h^2)}.2π(d2+h2)μ0​Id​.

Hence the total vertical field due to both wires is

Bwires=μ0Idπ(d2+h2),B_{\text{wires}}=\frac{\mu_0 I d}{\pi(d^2+h^2)},Bwires​=π(d2+h2)μ0​Id​,

provided their vertical components add.

  1. Which current directions make the wire-fields oppose the loop field?

The loop field at the point above the center is upward. Therefore, for net field to be zero, the combined field of the two wires must be downward.

Now check current directions:

  • If wire 1 carries current in direction PQPQPQ and wire 2 in direction SRSRSR, then by the right-hand rule, the vertical components of field at the point above the center are both in the same downward direction, so they add and can cancel the loop field.
  • If the currents are in directions PQPQPQ and RSRSRS, then the vertical components from the two wires are opposite and tend to cancel each other, so they cannot balance the loop field effectively.

Therefore the correct current directions must be:

wire 1: PQ,wire 2: SR.\text{wire 1: } PQ, \qquad \text{wire 2: } SR.wire 1: PQ,wire 2: SR.

So only B and C remain possible.

  1. Equation for zero net field

Set the loop field equal to the total field from the two wires:

μ0Ia22(a2+h2)3/2=μ0Idπ(d2+h2).\frac{\mu_0 I a^2}{2(a^2+h^2)^{3/2}}=\frac{\mu_0 I d}{\pi(d^2+h^2)}.2(a2+h2)3/2μ0​Ia2​=π(d2+h2)μ0​Id​.

Cancel μ0I\mu_0 Iμ0​I:

a22(a2+h2)3/2=dπ(d2+h2).\frac{a^2}{2(a^2+h^2)^{3/2}}=\frac{d}{\pi(d^2+h^2)}.2(a2+h2)3/2a2​=π(d2+h2)d​.

Given d≈ad\approx ad≈a, put d=ad=ad=a for estimation:

a22(a2+h2)3/2=aπ(a2+h2).\frac{a^2}{2(a^2+h^2)^{3/2}}=\frac{a}{\pi(a^2+h^2)}.2(a2+h2)3/2a2​=π(a2+h2)a​.

Multiply both sides by 2(a2+h2)3/22(a^2+h^2)^{3/2}2(a2+h2)3/2:

a2=2aπa2+h2.a^2 = \frac{2a}{\pi}\sqrt{a^2+h^2}.a2=π2a​a2+h2​.

So

a2+h2=πa2.\sqrt{a^2+h^2}=\frac{\pi a}{2}.a2+h2​=2πa​.

Squaring,

a2+h2=π2a24.a^2+h^2=\frac{\pi^2 a^2}{4}.a2+h2=4π2a2​.

Thus

h2=a2(π24−1).h^2=a^2\left(\frac{\pi^2}{4}-1\right).h2=a2(4π2​−1).

Hence

h=aπ24−1.h=a\sqrt{\frac{\pi^2}{4}-1}.h=a4π2​−1​.

Now,

π24≈9.874≈2.47,\frac{\pi^2}{4}\approx \frac{9.87}{4}\approx 2.47,4π2​≈49.87​≈2.47,

so

h≈a2.47−1=a1.47≈1.21a.h\approx a\sqrt{2.47-1}=a\sqrt{1.47}\approx 1.21a.h≈a2.47−1​=a1.47​≈1.21a.

Therefore,

h≈1.2a.h\approx 1.2a.h≈1.2a.
  1. Final option check
  • A: wrong current directions and wrong height.
  • B: correct current directions, but wrong height.
  • C: correct current directions and h≈1.2ah\approx 1.2ah≈1.2a.
  • D: wrong current directions.

Therefore the correct option is

C\boxed{\text{C}}C​
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