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Magnetics question

2025 · 2 Apr · Shift 1 · Q56
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Magnetics question

2025 · 2 Apr · Shift 1 · Q56

JEE MainPhysicsMagneticsMCQ+4 / −1
Let B1B_1B1​ be the magnitude of magnetic field at center of a circular coil of radius RRR carrying current I. Let B2\mathrm{B}_2B2​ be the magnitude of magnetic field at an axial distance 'xxx' from the center. For x:R=3:4, B2 B1x: \mathrm{R}=3: 4, \frac{\mathrm{~B}_2}{\mathrm{~B}_1}x:R=3:4, B1​ B2​​ is :
  1. A
    64:12564: 12564:125
  2. B
    25:1625: 1625:16
  3. C
    4:54: 54:5
  4. D
    16:2516: 2516:25
View written solutionFree

Correct answer: A

  1. Magnetic field at the center of a circular coil

For a circular loop of radius RRR carrying current III, the magnetic field at the center is

B1=μ0I2RB_1=\frac{\mu_0 I}{2R}B1​=2Rμ0​I​
  1. Magnetic field on the axis at distance xxx

At a point on the axis of the loop, at distance xxx from the center, the magnetic field is

B2=μ0IR22(R2+x2)3/2B_2=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}B2​=2(R2+x2)3/2μ0​IR2​
  1. Find the ratio B2B1\dfrac{B_2}{B_1}B1​B2​​

Divide B2B_2B2​ by B1B_1B1​:

B2B1=μ0IR22(R2+x2)3/2μ0I2R\frac{B_2}{B_1}= \frac{\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}}{\dfrac{\mu_0 I}{2R}}B1​B2​​=2Rμ0​I​2(R2+x2)3/2μ0​IR2​​

Cancelling common terms:

B2B1=R3(R2+x2)3/2\frac{B_2}{B_1}=\frac{R^3}{(R^2+x^2)^{3/2}}B1​B2​​=(R2+x2)3/2R3​
  1. Use the given ratio x:R=3:4x:R=3:4x:R=3:4

So let

x=3k,R=4kx=3k, \quad R=4kx=3k,R=4k

Then

R2+x2=(4k)2+(3k)2=16k2+9k2=25k2R^2+x^2=(4k)^2+(3k)^2=16k^2+9k^2=25k^2R2+x2=(4k)2+(3k)2=16k2+9k2=25k2

Hence,

(R2+x2)3/2=(25k2)3/2=125k3(R^2+x^2)^{3/2}=(25k^2)^{3/2}=125k^3(R2+x2)3/2=(25k2)3/2=125k3

Also,

R3=(4k)3=64k3R^3=(4k)^3=64k^3R3=(4k)3=64k3

Therefore,

B2B1=64k3125k3=64125\frac{B_2}{B_1}=\frac{64k^3}{125k^3}=\frac{64}{125}B1​B2​​=125k364k3​=12564​
  1. Match with options
B2B1=64:125\frac{B_2}{B_1}=64:125B1​B2​​=64:125

So the correct option is A.

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