A long solenoid is formed by winding 40 turns/cm. The current required to produce a magnetic field of 22 mT inside the solenoid will be
(Take $μ_0 = 4 π × 10^{-7} T m A^{-1}$)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.4 A **
Given:
Number of turns per cm = 40 turns/cm
Magnetic field, $B = 22\ \text{mT} = 22 \times 10^{-3}\ \text{T}$
$\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$
For a long solenoid: $B = \mu_0 n I$
where $n$ = number of turns per meter = $40 \times 100 = 4000\ \text{turns/m}$
Substituting,
$22 \times 10^{-3} = 4\pi \times 10^{-7} \times 4000 \times I$
$I = \frac{22 \times 10^{-3}}{4\pi \times 10^{-7} \times 4000}$
$I = \frac{22 \times 10^{-3}}{1.6\pi \times 10^{-3}} = \frac{22}{1.6\pi}$
$I \approx 4.38\ \text{A}$
Hence, the required current is approximately 4.4 A.