Question: 31: A long solenoid of radius 1 mm has 100 turns per mm . If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is
(1) 12.56 \times 10^{-4} \mathrm{~T}
(2) 6.28 \times 10^{-4} \mathrm{~T}
(3) 6.28 \times 10^{-2} \mathrm{~T}
(4) 12.56 \times 10^{-2} \mathrm{~T}
Answer: Option (4)
Explanation:
For a long solenoid, the magnetic field at its centre is given by the formula B = \mu_0 n I,
where \mu_0 is the permeability of free space, n is the number of turns per unit length, and I is the current.
Given that the solenoid has 100 turns per mm, converting this into turns per metre gives
n = 100 \times 10^3 \ \text{turns m}^{-1}.
The value of permeability of free space is \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T\,m\,A^{-1}}, and the current is I = 1 \ \mathrm{A}.
Substituting the values in the formula, we get
B = 4\pi \times 10^{-7} \times 100 \times 10^3 \times 1.
Simplifying, B = 4\pi \times 10^{-2} \ \mathrm{T}.
Since \pi \approx 3.14,
The magnetic field becomes
B = 12.56 \times 10^{-2} \ \mathrm{T}.
Hence, the correct answer is Option (4).