A solenoid of length 30 cm and 3000 turns carries a current of 10 A. The value of the magnetic field ($\vec B$) at the centre of solenoid is:
(Given $μ_0 = 4π × 10^{-7} Wb/A-m$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $0.13\, Wb/m^2$
Magnetic field inside a solenoid: $B = \mu_0 n I$
Length of solenoid, $l = 30 \, \text{cm} = 0.30 \, \text{m}$
Number of turns, $N = 3000$
Turns per unit length, $n = \frac{N}{l} = \frac{3000}{0.30} = 10000 \, \text{turns/m}$
Current, $I = 10 \, \text{A}$
Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}$
$B = \mu_0 n I = (4\pi \times 10^{-7})(10000)(10)$
$B = 4\pi \times 10^{-7} \times 10^5$
$B = 4\pi \times 10^{-2} \, \text{T}$
$B \approx 0.126 \, \text{T}$
Answer: $B = 0.13 \, \text{T}$