Question

A long solenoid of $50\,cm$ length having $100\,turns$ carries a current of $2.5\,A$. The magnetic field at the centre of the solenoid is : $(\mu_o = 4\pi \times 10^{-7} T\,m\,A^{-1})$

• A. $6.28 \times 10^{-4}T$
• B. $3.14 \times 10^{-4}T$
• C. $6.28 \times 10^{-5}T$
• D. $3.14 \times 10^{-5}T$

Answer 1

The Correct Option is A

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

\(B = \mu_0 \cdot n \cdot I\)

where:

• \(\mu_0\) is the permeability of free space, given as \(4\pi \times 10^{-7} \, T \cdot m \cdot A^{-1}\)
• \(n\) is the number of turns per unit length of the solenoid
• \(I\) is the current passing through the solenoid

Given data:

• Length of the solenoid \(L = 50 \, cm = 0.5 \, m\)
• Number of turns \(N = 100\)
• Current \(I = 2.5 \, A\)

First, we calculate \(n\):

\(n = \frac{N}{L} = \frac{100}{0.5} = 200 \, \text{turns per meter}\)

Now, substituting these values into the formula for the magnetic field, \(B\):

\(B = (4\pi \times 10^{-7} \, T \cdot m \cdot A^{-1}) \cdot 200 \, m^{-1} \cdot 2.5 \, A\)

Calculating the magnetic field:

\(B = 4\pi \times 10^{-7} \cdot 200 \cdot 2.5\)

\(B = 4\pi \times 500 \times 10^{-7}\)

\(B = 2000\pi \times 10^{-7}\)

Approximate the value of \(\pi \approx 3.14\):

\(B = 2000 \times 3.14 \times 10^{-7} \, T\)

\(B = 6280 \times 10^{-7} \, T\)

\(B = 6.28 \times 10^{-4} \, T\)

Thus, the magnetic field at the centre of the solenoid is \(6.28 \times 10^{-4} \, T\), which matches the correct option provided.