Question

The flow of current of \(2\) A through a straight solenoid of length \(2\) m produces a magnetic field of \(2\pi \times 10^{-4}\) T at its centre. Then the number of turns in the solenoid is

• A. $600$
• B. $500$
• C. $3500$
• D. $5000$
• E. $700$

Hint:

For solenoids: - $B = \mu_0 n I$ - Always convert $n = \frac{N}{L}$

Answer 1

The Correct Option is B

Concept: Magnetic field inside a solenoid: \[ B = \mu_0 n I \] where $n = \frac{N}{L}$.

Step 1: Substitute $n = \frac{N}{L}$.
\[ B = \mu_0 \frac{N}{L} I \]

Step 2: Solve for $N$.
\[ N = \frac{BL}{\mu_0 I} \]

Step 3: Substitute values.
\[ B = 2\pi \times 10^{-4},\quad L = 2,\quad I = 2,\quad \mu_0 = 4\pi \times 10^{-7} \] \[ N = \frac{(2\pi \times 10^{-4}) \times 2}{(4\pi \times 10^{-7}) \times 2} \]

Step 4: Simplify.
\[ N = \frac{4\pi \times 10^{-4}}{8\pi \times 10^{-7}} = \frac{4}{8} \times 10^{3} = 0.5 \times 10^3 = 500 \]