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Magnetics question

2002 · Shift 0 · Q123
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Magnetics question

2002 · Shift 0 · Q123

JEE MainPhysicsMagneticsMCQ+4 / −1
If in a circular coil AAA of radius R,R,R, current III is flowing and in another coil BBB of radius 2R2R2R a current 2I2I2I is flowing, then the ratio of the magnetic fields BA{B_A}BA​ and BB{B_B}BB​, produced by them will be
  1. A
    111
  2. B
    222
  3. C
    1/21/21/2
  4. D
    444
View written solutionFree

Correct answer: A

  1. Magnetic field at the center of a circular coil

For a single circular coil of radius rrr carrying current iii, the magnetic field at the center is

B=μ0i2rB = \frac{\mu_0 i}{2r}B=2rμ0​i​

  1. Field due to coil AAA

For coil AAA:

  • radius =R= R=R
  • current =I= I=I

So,

BA=μ0I2RB_A = \frac{\mu_0 I}{2R}BA​=2Rμ0​I​

  1. Field due to coil BBB

For coil BBB:

  • radius =2R= 2R=2R
  • current =2I= 2I=2I

So,

BB=μ0(2I)2(2R)=μ02I4R=μ0I2RB_B = \frac{\mu_0 (2I)}{2(2R)} = \frac{\mu_0 2I}{4R} = \frac{\mu_0 I}{2R}BB​=2(2R)μ0​(2I)​=4Rμ0​2I​=2Rμ0​I​

  1. Find the ratio

BABB=μ0I2Rμ0I2R=1\frac{B_A}{B_B} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{2R}} = 1BB​BA​​=2Rμ0​I​2Rμ0​I​​=1

So,

BA:BB=1:1B_A : B_B = 1 : 1BA​:BB​=1:1

  1. Option check
  • A: 111 ✅
  • B: 222 ❌
  • C: 1/21/21/2 ❌
  • D: 444 ❌

Hence, the correct answer is A.

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