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Moving Charges and Magnetism question

2020 · Q132
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Moving Charges and Magnetism question

2020 · Q132

NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1
A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is :
(μ0=4π×10−7TmA−1)\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}Tm{{A}^{-1}}} \right)(μ0​=4π×10−7TmA−1)
  1. A
    3.14 ×10−4T{ \times {{10}^{ - 4}}T}×10−4T
  2. B
    6.28 ×10−5T{ \times {{10}^{ - 5}}T}×10−5T
  3. C
    3.14 ×10−5T{ \times {{10}^{ - 5}}T}×10−5T
  4. D
    6.28 ×10−4T{ \times {{10}^{ - 4}}T}×10−4T
View written solutionFree

Correct answer: D

Magnetic field at centre of solenoid = μ0nI{\mu _0}nIμ0​nI
n=NL=10050×10−2n = {N \over L} = {{100} \over {50 \times {{10}^{ - 2}}}}n=LN​=50×10−2100​
= 200 turns/m
I = 25A
Now, substituting the values,
B = 4π\pi π ×\times× 10-7 ×\times× 200 ×\times× 2.5
= 6.28 ×\times× 10-4 T

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