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Magnetics question

2021 · 17 Mar · Shift 1 · Q52
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  5. /2021 · 17 Mar · Shift 1 · Q52

Magnetics question

2021 · 17 Mar · Shift 1 · Q52

JEE MainPhysicsMagneticsMCQ+4 / −1
A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is : (permeability of free space = 4 π×\pi\timesπ× 10 −-− 7 H/m)
  1. A
    π\piπ T
  2. B
    2 ×\times× 10 −-− 3 π\piπ T
  3. C
    10 −-− 4 π\piπ T
  4. D
    π5{\pi \over 5}5π​ T
View written solutionFree

Correct answer: A

  1. Magnetic flux density inside a solenoid

For a long solenoid with a magnetic core,

B=μnIB = \mu n IB=μnI

where:

  • μ=μ0μr\mu = \mu_0 \mu_rμ=μ0​μr​
  • μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7}\,\text{H/m}μ0​=4π×10−7H/m
  • μr=500\mu_r = 500μr​=500
  • n=1000 turns/mn = 1000\,\text{turns/m}n=1000turns/m
  • I=5 AI = 5\,\text{A}I=5A
  1. Substitute μ\muμ

μ=μ0μr=(4π×10−7)(500)\mu = \mu_0 \mu_r = (4\pi \times 10^{-7})(500)μ=μ0​μr​=(4π×10−7)(500)

  1. Now compute BBB

B=(4π×10−7)(500)(1000)(5)B = (4\pi \times 10^{-7})(500)(1000)(5)B=(4π×10−7)(500)(1000)(5)

First multiply the numerical factors:

500×1000×5=2.5×106500 \times 1000 \times 5 = 2.5 \times 10^6500×1000×5=2.5×106

So,

B=4π×10−7×2.5×106B = 4\pi \times 10^{-7} \times 2.5 \times 10^6B=4π×10−7×2.5×106

B=10π×10−1B = 10\pi \times 10^{-1}B=10π×10−1

B=π TB = \pi\,\text{T}B=πT

  1. Match with the options

Option A is:

π T\pi\,\text{T}πT

So the correct answer is A.

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