Back to Directory
NEET PHYSICSEasy

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 T m A−1)(\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1})

A

3.4×10−4 T3.4 \times 10^{-4} \text{ T}

B

6.28×10−5 T6.28 \times 10^{-5} \text{ T}

C

3.14×10−5 T3.14 \times 10^{-5} \text{ T}

D

6.28×10−4 T6.28 \times 10^{-4} \text{ T}

Step-by-Step Solution

  1. Formula: The magnetic field BB inside a long solenoid is given by the formula B=μ0nIB = \mu_0 n I, where μ0\mu_0 is the permeability of free space, nn is the number of turns per unit length, and II is the current .
  2. Calculate nn: The number of turns per unit length nn is calculated as n=NLn = \frac{N}{L}. Given N=100N = 100 turns and L=50 cm=0.5 mL = 50 \text{ cm} = 0.5 \text{ m}. n=1000.5=200 turns/mn = \frac{100}{0.5} = 200 \text{ turns/m}
  3. Substitute Values: B=(4π×10−7 T m A−1)×(200 m−1)×(2.5 A)B = (4\pi \times 10^{-7} \text{ T m A}^{-1}) \times (200 \text{ m}^{-1}) \times (2.5 \text{ A}) B=4×3.14×10−7×500B = 4 \times 3.14 \times 10^{-7} \times 500 B=6280×10−7 TB = 6280 \times 10^{-7} \text{ T} B=6.28×10−4 TB = 6.28 \times 10^{-4} \text{ T}
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started