Question

A solenoid has 1000 turns per meter and carries a current of \( \frac{7}{\pi} \) A. The magnetic field inside the solenoid is

Hint:

Be careful to distinguish between total turns '$N$' and turns per unit length '$n$'. The formula uses $n = N/L$. The problem directly gives '$n$' (1000 turns per meter).

Answer 1

Step 1: Understanding the Concept:
An ideal solenoid is a long coil of wire. When current flows through it, it generates a nearly uniform and strong magnetic field concentrated strictly along its interior central axis.

Step 2: Key Formula or Approach:
The magnitude of the magnetic field ($B$) well inside a long solenoid is given by Ampere's Law:
\[ B = \mu_0 n I \]
Where:
$\mu_0$ = permeability of free space = $4\pi \times 10^{-7}$ T$\cdot$m/A
$n$ = number of turns per unit length (turns/meter)
$I$ = current in amperes

Step 3: Detailed Explanation:
Given values:
Turn density, $n = 1000$ turns/m = $10^3$ m$^{-1}$
Current, $I = \frac{7}{\pi}$ A
Substitute these into the formula:
\[ B = (4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}) \times (10^3 \text{ m}^{-1}) \times \left(\frac{7}{\pi} \text{ A}\right) \]
Notice how the $\pi$ terms cleanly cancel out:
\[ B = 4 \times 10^{-7} \times 10^3 \times 7 \]
Group the numbers and powers of 10:
\[ B = (4 \times 7) \times (10^{-7} \times 10^3) \]
\[ B = 28 \times 10^{-4} \text{ T} \]
To write this in standard scientific notation:
\[ B = 2.8 \times 10^{-3} \text{ T} \]

Step 4: Final Answer:
The magnetic field is $2.8 \times 10^{-3}$ T.