A solenoid of length 10 cm and radius 1 cm contains 200 turns and carries a current of 10 A. Find the magnetic field at a point on the axis at a distance of 10 cm from the centre.

Steps to Solve

1. Identify the given values
   We have the following information about the solenoid:

• Length of solenoid, $L = 10 \text{ cm} = 0.1 \text{ m}$
• Radius of solenoid, $r = 1 \text{ cm} = 0.01 \text{ m}$
• Number of turns, $N = 200$
• Current, $I = 10 \text{ A}$

2. Calculate the magnetic field inside the solenoid
   For a long solenoid, the magnetic field strength $B$ inside the solenoid is given by the formula:
   $$ B = \mu_0 \cdot n \cdot I $$
   where $\mu_0 = 4\pi \times 10^{-7} , \text{T m/A}$ (the permeability of free space) and ( n ) is the number of turns per unit length, calculated as:
   $$ n = \frac{N}{L} = \frac{200}{0.1} = 2000 , \text{turns/m} $$
   Now we can substitute these values into the formula for $B$:
   $$ B = (4\pi \times 10^{-7}) \cdot (2000) \cdot (10) $$

3. Calculate the value of ( B )
   Now we simplify the expression for $B$:
   $$ B = 4\pi \times 10^{-7} \cdot 20000 $$
   $$ B = 8\pi \times 10^{-3} , \text{T} $$

4. Evaluate ( B )
   Calculate the numerical value of $B$:
   Using $\pi \approx 3.14$:
   $$ B \approx 8 \times 3.14 \times 10^{-3} \approx 0.02512 , \text{T} $$
   Thus, the magnetic field at the given point on the axis of the solenoid is approximately $0.02512 , \text{T}$ or $25.12 , \text{mT}$.