Verified Solution • 2025–26 Session moderate

QNEET-2020-Ph-702: NEET 2020 Physics - Class 12 NEET-2020-Ph-702 Solution: A long solenoid of 50 cm length having 100 turns carries a…

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×10−7 TmA−1\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) 3.14×10−4 T3.14 \times 10^{-4}\ \text{T} 6.28×10−5 T6.28 \times 10^{-5}\ \text{T} 3.14×10−5 T3.14 \times 10^{-5}\ \text{T} 6.28×10−4 T6.28 \times 10^{-4}\ \text{T}

Multiple Choice Options

  • A 3.14 ×10−4T{ \times {{10}^{ - 4}}T} _ 10 _ 4 T
  • B 6.28 ×10−5T{ \times {{10}^{ - 5}}T} _ 10 _ 5 T
  • C 3.14 ×10−5T{ \times {{10}^{ - 5}}T} _ 10 _ 5 T
  • D 6.28 ×10−4T{ \times {{10}^{ - 4}}T} _ 10 _ 4 T

Step-by-Step Verified Solution

Magnetic field at centre of solenoid = _ 0 n I n = N L = 100 50 _ 10 _ 2 = 200 turns/m I = 25A Now, substituting the values, B = 4 � _ 10 -7 _ 200 _ 2.5 = 6.28 _ 10 -4 T

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