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Magnetics question

2024 · 5 Apr · Shift 2 · Q90
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  5. /2024 · 5 Apr · Shift 2 · Q90

Magnetics question

2024 · 5 Apr · Shift 2 · Q90

JEE MainPhysicsMagneticsNumerical+4 / −1
A solenoid of length 0.5 m0.5 \mathrm{~m}0.5 m has a radius of 1 cm1 \mathrm{~cm}1 cm and is made up of 'm\mathrm{m}m' number of turns. It carries a current of 5 A5 \mathrm{~A}5 A. If the magnitude of the magnetic field inside the solenoid is 6.28×10−3 T6.28 \times 10^{-3} \mathrm{~T}6.28×10−3 T then the value of m\mathrm{m}m is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 500

  1. For a long solenoid, the magnetic field inside is

B=μ0nIB=\mu_0 n IB=μ0​nI

where:

  • B=6.28×10−3 TB = 6.28\times 10^{-3}\,\text{T}B=6.28×10−3T
  • μ0=4π×10−7 T m/A\mu_0 = 4\pi\times 10^{-7}\,\text{T m/A}μ0​=4π×10−7T m/A
  • n=mLn = \dfrac{m}{L}n=Lm​ is the number of turns per unit length
  • I=5 AI = 5\,\text{A}I=5A
  • L=0.5 mL = 0.5\,\text{m}L=0.5m
  1. Substitute n=mLn=\dfrac{m}{L}n=Lm​:

B=μ0mLIB=\mu_0 \frac{m}{L} IB=μ0​Lm​I

So,

m=BLμ0Im=\frac{BL}{\mu_0 I}m=μ0​IBL​

  1. Put the given values:

m=(6.28×10−3)(0.5)(4π×10−7)(5)m=\frac{(6.28\times 10^{-3})(0.5)}{(4\pi\times 10^{-7})(5)}m=(4π×10−7)(5)(6.28×10−3)(0.5)​

  1. Simplify numerator:

(6.28×10−3)(0.5)=3.14×10−3(6.28\times 10^{-3})(0.5)=3.14\times 10^{-3}(6.28×10−3)(0.5)=3.14×10−3

So,

m=3.14×10−320π×10−7m=\frac{3.14\times 10^{-3}}{20\pi\times 10^{-7}}m=20π×10−73.14×10−3​

Since 20π×10−7=2π×10−620\pi\times 10^{-7}=2\pi\times 10^{-6}20π×10−7=2π×10−6,

m=3.14×10−32π×10−6m=\frac{3.14\times 10^{-3}}{2\pi\times 10^{-6}}m=2π×10−63.14×10−3​

Using π≈3.14\pi\approx 3.14π≈3.14,

m=3.14×10−36.28×10−6=0.5×103=500m=\frac{3.14\times 10^{-3}}{6.28\times 10^{-6}}=0.5\times 10^3=500m=6.28×10−63.14×10−3​=0.5×103=500

  1. Therefore, the number of turns is

500\boxed{500}500​

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