A solenoid of length $50~\mathrm{cm}$ and a radius of cross-section $1\mathrm{cm}$ has 1000 turns of wire wound over it. If the current carried is 5A, the magnetic field on its axis, near the centre of the solenoid is approximately (permeability of free space $\mu_{0} = 4\pi \times 10^{-7}\mathrm{T}\cdot \mathrm{m / A}$)

Question

The pattern of the magnetic field produced inside a current carrying solenoid is :

Answer 1

To determine the magnetic field at the center of a solenoid, we use the formula for the magnetic field inside a solenoid:

$B = \mu_{0} n I$

where:

$B$ is the magnetic field at the center of the solenoid.
$\mu_{0} = 4\pi \times 10^{-7} \, \mathrm{T \, m/A}$ is the permeability of free space.
$n$ is the number of turns per unit length of the solenoid.
$I$ is the current flowing through the solenoid.

Given:

First, we calculate the number of turns per unit length $n$:

$n = \frac{N}{L} = \frac{1000}{0.5} = 2000 \, \text{turns/m}$

Now, substitute these values into the formula for the magnetic field:

$B = \mu_{0} n I = 4\pi \times 10^{-7} \times 2000 \times 5$

Calculate $B$:

$B = 4\pi \times 10^{-7} \times 10000 = 4\pi \times 10^{-3} \, \mathrm{T}$

Using the approximation $\pi \approx 3.14$:

$B = 4 \times 3.14 \times 10^{-3} \approx 12.56 \times 10^{-3} \, \mathrm{T}$

$B = 1.26 \times 10^{-2} \, \mathrm{T}$

Therefore, the magnetic field on the axis near the center of the solenoid is approximately $1.26\times 10^{-2} \mathrm{T}$.

The correct answer is: 1.26 × 10^-2 T

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