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Magnetics question

2021 · 26 Aug · Shift 1 · Q43
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  5. /2021 · 26 Aug · Shift 1 · Q43

Magnetics question

2021 · 26 Aug · Shift 1 · Q43

JEE MainPhysicsMagneticsMCQ+4 / −1
The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is : (Take r < a)
  1. A
    32a2r2{3 \over 2}{{{a^2}} \over {{r^2}}}23​r2a2​
  2. B
    23a2r2{2 \over 3}{{{a^2}} \over {{r^2}}}32​r2a2​
  3. C
    23r2a2{2 \over 3}{{{r^2}} \over {{a^2}}}32​a2r2​
  4. D
    32r2a2{3 \over 2}{{{r^2}} \over {{a^2}}}23​a2r2​
View written solutionFree

Correct answer: D

  1. Magnetic field on the axis of a circular coil

For a circular coil of radius aaa carrying current III, the magnetic field at a point on its axis at distance rrr from the centre is

Br=μ0Ia22(a2+r2)3/2.B_r=\frac{\mu_0 I a^2}{2(a^2+r^2)^{3/2}}.Br​=2(a2+r2)3/2μ0​Ia2​.

At the centre of the coil (r=0)(r=0)(r=0),

B0=μ0I2a.B_0=\frac{\mu_0 I}{2a}.B0​=2aμ0​I​.
  1. Find the ratio BrB0\dfrac{B_r}{B_0}B0​Br​​
BrB0=μ0Ia22(a2+r2)3/2⋅2aμ0I=a3(a2+r2)3/2.\frac{B_r}{B_0}= \frac{\mu_0 I a^2}{2(a^2+r^2)^{3/2}}\cdot \frac{2a}{\mu_0 I} =\frac{a^3}{(a^2+r^2)^{3/2}}.B0​Br​​=2(a2+r2)3/2μ0​Ia2​⋅μ0​I2a​=(a2+r2)3/2a3​.

So,

BrB0=(a2a2+r2)3/2=(1+r2a2)−3/2.\frac{B_r}{B_0}=\left(\frac{a^2}{a^2+r^2}\right)^{3/2} =\left(1+\frac{r^2}{a^2}\right)^{-3/2}.B0​Br​​=(a2+r2a2​)3/2=(1+a2r2​)−3/2.
  1. Use binomial expansion since r<ar<ar<a

Because r2a2<1\dfrac{r^2}{a^2}<1a2r2​<1, we expand:

(1+x)−3/2≈1−32x(1+x)^{-3/2}\approx 1-\frac{3}{2}x(1+x)−3/2≈1−23​x

for small xxx.

Here,

x=r2a2.x=\frac{r^2}{a^2}.x=a2r2​.

Therefore,

BrB0≈1−32r2a2.\frac{B_r}{B_0}\approx 1-\frac{3}{2}\frac{r^2}{a^2}.B0​Br​​≈1−23​a2r2​.
  1. Fractional change in magnetic field

Fractional change from centre value is

B0−BrB0=1−BrB0.\frac{B_0-B_r}{B_0}=1-\frac{B_r}{B_0}.B0​B0​−Br​​=1−B0​Br​​.

Thus,

B0−BrB0≈1−(1−32r2a2)=32r2a2.\frac{B_0-B_r}{B_0} \approx 1-\left(1-\frac{3}{2}\frac{r^2}{a^2}\right) =\frac{3}{2}\frac{r^2}{a^2}.B0​B0​−Br​​≈1−(1−23​a2r2​)=23​a2r2​.
  1. Match with options
32r2a2\boxed{\frac{3}{2}\frac{r^2}{a^2}}23​a2r2​​

So the correct option is D.

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