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Electromagnetic Induction question

2013 · Shift 0 · Q59
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  5. /2013 · Shift 0 · Q59

Electromagnetic Induction question

2013 · Shift 0 · Q59

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A circular loop of radius 0.3cm0.3cm0.3cm lies center of the small loop is on the axis of the bigger loop. The distance between their centers is 15cm.15cm.15cm. If a current of 2.0A2.0A2.0A flows through the smaller loop, than the flux linked with bigger loop is
  1. A
    9.1×10−11 9.1 \times {10^{ - 11}}\,9.1×10−11 weber
  2. B
    6×10−11 6 \times {10^{ - 11}}\,6×10−11 weber
  3. C
    3.3×10−11 3.3 \times {10^{ - 11}}\,3.3×10−11 weber
  4. D
    6.6×10−9 6.6 \times {10^{ - 9}}\,6.6×10−9 weber
View written solutionFree

Correct answer: A

  1. Interpretation of the question

A small circular loop carries current I=2.0 AI=2.0\,\text{A}I=2.0A. We need the magnetic flux linked with a bigger loop whose center lies on the axis of the small loop.

Given:

  • Radius of small loop: r=0.3 cm=3×10−3 mr=0.3\,\text{cm}=3\times 10^{-3}\,\text{m}r=0.3cm=3×10−3m
  • Distance between centers: x=15 cm=0.15 mx=15\,\text{cm}=0.15\,\text{m}x=15cm=0.15m

The bigger loop radius is not explicitly visible in the statement, but from the options and standard form of this problem, it is R=0.2 mR=0.2\,\text{m}R=0.2m.

  1. Magnetic field on the axis of a circular loop

The magnetic field at a point on the axis of a circular loop is

B=μ0Ir22(x2+r2)3/2B=\frac{\mu_0 I r^2}{2(x^2+r^2)^{3/2}}B=2(x2+r2)3/2μ0​Ir2​

Since x≫rx \gg rx≫r, we may also use the exact expression directly.

Substitute values:

μ0=4π×10−7 H/m\mu_0=4\pi\times 10^{-7}\,\text{H/m}μ0​=4π×10−7H/m r2=(3×10−3)2=9×10−6r^2=(3\times 10^{-3})^2=9\times 10^{-6}r2=(3×10−3)2=9×10−6 x2+r2=(0.15)2+9×10−6=0.0225+0.000009=0.022509x^2+r^2=(0.15)^2+9\times 10^{-6}=0.0225+0.000009=0.022509x2+r2=(0.15)2+9×10−6=0.0225+0.000009=0.022509 (x2+r2)3/2≈(0.022509)3/2≈0.003377(x^2+r^2)^{3/2}\approx (0.022509)^{3/2}\approx 0.003377(x2+r2)3/2≈(0.022509)3/2≈0.003377

Therefore,

B=(4π×10−7)(2)(9×10−6)2(0.003377)B=\frac{(4\pi\times 10^{-7})(2)(9\times 10^{-6})}{2(0.003377)}B=2(0.003377)(4π×10−7)(2)(9×10−6)​

Cancel the factor 222:

B=4π×10−7×9×10−60.003377B=\frac{4\pi\times 10^{-7}\times 9\times 10^{-6}}{0.003377}B=0.0033774π×10−7×9×10−6​ B≈3.35×10−9 TB\approx 3.35\times 10^{-9}\,\text{T}B≈3.35×10−9T
  1. Flux linked with the bigger loop

If the bigger loop is sufficiently small compared to the distance scale over which the field changes, flux is

Φ=BA\Phi = BAΦ=BA

Area of bigger loop:

A=πR2=π(0.2)2=0.04π m2A=\pi R^2=\pi(0.2)^2=0.04\pi\,\text{m}^2A=πR2=π(0.2)2=0.04πm2

So,

Φ=(3.35×10−9)(0.04π)\Phi = (3.35\times 10^{-9})(0.04\pi)Φ=(3.35×10−9)(0.04π) Φ≈4.2×10−10 Wb\Phi \approx 4.2\times 10^{-10}\,\text{Wb}Φ≈4.2×10−10Wb

This does not match any option, so let us use the mutual inductance approximation for two coaxial loops with one very small loop and one larger loop far away. In such problems, the flux linked with the bigger loop due to the small loop is more conveniently obtained using reciprocity:

M=μ0πr2R22(x2+R2)3/2M = \frac{\mu_0 \pi r^2 R^2}{2(x^2+R^2)^{3/2}}M=2(x2+R2)3/2μ0​πr2R2​

Then

Φ=MI\Phi = MIΦ=MI

Substitute:

  • r=3×10−3 mr=3\times 10^{-3}\,\text{m}r=3×10−3m
  • R=0.2 mR=0.2\,\text{m}R=0.2m
  • x=0.15 mx=0.15\,\text{m}x=0.15m
  • I=2 AI=2\,\text{A}I=2A
Φ=μ0πr2R2I2(x2+R2)3/2\Phi = \frac{\mu_0 \pi r^2 R^2 I}{2(x^2+R^2)^{3/2}}Φ=2(x2+R2)3/2μ0​πr2R2I​ =(4π×10−7)π(9×10−6)(0.04)(2)2(0.152+0.22)3/2=\frac{(4\pi\times10^{-7})\pi(9\times10^{-6})(0.04)(2)}{2(0.15^2+0.2^2)^{3/2}}=2(0.152+0.22)3/2(4π×10−7)π(9×10−6)(0.04)(2)​

Now,

0.152+0.22=0.0225+0.04=0.06250.15^2+0.2^2=0.0225+0.04=0.06250.152+0.22=0.0225+0.04=0.0625 (0.0625)3/2=(0.25)3=0.015625(0.0625)^{3/2}=(0.25)^3=0.015625(0.0625)3/2=(0.25)3=0.015625

Thus,

Φ=(4π2×10−7)(9×10−6)(0.04)0.015625\Phi=\frac{(4\pi^2\times10^{-7})(9\times10^{-6})(0.04)}{0.015625}Φ=0.015625(4π2×10−7)(9×10−6)(0.04)​ Φ≈9.1×10−11 Wb\Phi \approx 9.1\times10^{-11}\,\text{Wb}Φ≈9.1×10−11Wb
  1. Match with options

This matches:

9.1×10−11 Wb\boxed{9.1\times10^{-11}\,\text{Wb}}9.1×10−11Wb​

So the correct option is A.

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