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Magnetics question

2022 · 26 Jul · Shift 1 · Q52
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  5. /2022 · 26 Jul · Shift 1 · Q52

Magnetics question

2022 · 26 Jul · Shift 1 · Q52

JEE MainPhysicsMagneticsMCQ+4 / −1
BX\mathrm{B}_{X}BX​ and BY\mathrm{B}_{\mathrm{Y}}BY​ are the magnetic fields at the centre of two coils X\mathrm{X}X and Y\mathrm{Y}Y respectively each carrying equal current. If coil XXX has 200 turns and 20 cm20 \mathrm{~cm}20 cm radius and coil YYY has 400 turns and 20 cm20 \mathrm{~cm}20 cm radius, the ratio of BXB_{X}BX​ and BYB_{Y}BY​ is :
  1. A
    1 : 1
  2. B
    1 : 2
  3. C
    2 : 1
  4. D
    4 : 1
View written solutionFree

Correct answer: B

  1. Magnetic field at the center of a circular coil

For a coil of NNN turns, radius RRR, carrying current III, the magnetic field at the center is

B=μ0NI2RB = \frac{\mu_0 N I}{2R}B=2Rμ0​NI​

  1. Given data

For coil XXX:

  • NX=200N_X = 200NX​=200
  • RX=20 cmR_X = 20\text{ cm}RX​=20 cm

For coil YYY:

  • NY=400N_Y = 400NY​=400
  • RY=20 cmR_Y = 20\text{ cm}RY​=20 cm

Also, both coils carry the same current.

  1. Write expressions for BXB_XBX​ and BYB_YBY​

BX=μ0NXI2RX=μ0⋅200⋅I2⋅20 cmB_X = \frac{\mu_0 N_X I}{2R_X} = \frac{\mu_0 \cdot 200 \cdot I}{2 \cdot 20\text{ cm}}BX​=2RX​μ0​NX​I​=2⋅20 cmμ0​⋅200⋅I​

BY=μ0NYI2RY=μ0⋅400⋅I2⋅20 cmB_Y = \frac{\mu_0 N_Y I}{2R_Y} = \frac{\mu_0 \cdot 400 \cdot I}{2 \cdot 20\text{ cm}}BY​=2RY​μ0​NY​I​=2⋅20 cmμ0​⋅400⋅I​

  1. Find the ratio

Since current and radius are the same for both coils,

BXBY=NXNY=200400=12\frac{B_X}{B_Y} = \frac{N_X}{N_Y} = \frac{200}{400} = \frac{1}{2}BY​BX​​=NY​NX​​=400200​=21​

So,

BX:BY=1:2B_X : B_Y = 1 : 2BX​:BY​=1:2

  1. Match with options

The correct option is:

B: 1:21:21:2

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