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Magnetics question

2023 · 24 Jan · Shift 2 · Q47
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  5. /2023 · 24 Jan · Shift 2 · Q47

Magnetics question

2023 · 24 Jan · Shift 2 · Q47

JEE MainPhysicsMagneticsMCQ+4 / −1
A long solenoid is formed by winding 70 turns cm −1^{-1}−1. If 2.0 A current flows, then the magnetic field produced inside the solenoid is ‾\underline{\hspace{2cm}}​ (μ0=4π×10−7\mu_0=4\pi\times10^{-7}μ0​=4π×10−7 TmA −1^{-1}−1)
  1. A
    88×10−488\times10^{-4}88×10−4 T
  2. B
    1232×10−41232\times10^{-4}1232×10−4 T
  3. C
    176×10−4176\times10^{-4}176×10−4 T
  4. D
    352×10−4352\times10^{-4}352×10−4 T
View written solutionFree

Correct answer: C

  1. Formula for magnetic field inside a long solenoid

For a long solenoid,

B=μ0nIB = \mu_0 n IB=μ0​nI

where:

  • μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}μ0​=4π×10−7T m A−1
  • nnn = number of turns per metre
  • I=2.0 AI = 2.0\,\text{A}I=2.0A
  1. Convert turns per cm to turns per m

Given:

70 turns cm−170\,\text{turns cm}^{-1}70turns cm−1

Since 1 m=100 cm1\,\text{m} = 100\,\text{cm}1m=100cm,

n=70×100=7000 turns m−1n = 70 \times 100 = 7000\,\text{turns m}^{-1}n=70×100=7000turns m−1
  1. Substitute into the formula
B=(4π×10−7)(7000)(2)B = (4\pi \times 10^{-7})(7000)(2)B=(4π×10−7)(7000)(2) B=4π×10−7×14000B = 4\pi \times 10^{-7} \times 14000B=4π×10−7×14000 B=56π×10−4 TB = 56\pi \times 10^{-4}\,\text{T}B=56π×10−4T

Using π≈227\pi \approx \frac{22}{7}π≈722​,

B=56×227×10−4B = 56 \times \frac{22}{7} \times 10^{-4}B=56×722​×10−4 B=8×22×10−4B = 8 \times 22 \times 10^{-4}B=8×22×10−4 B=176×10−4 TB = 176 \times 10^{-4}\,\text{T}B=176×10−4T
  1. Match with the options

The correct option is:

C: 176×10−4 T\boxed{C:\ 176\times10^{-4}\,\text{T}}C: 176×10−4T​
  1. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.

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