Question

A solenoid of length 50 cm having 100 turns carries a current of 2.5 A. The magnetic field at one end of solenoid is

• A. 1.57 x 10^-4 T
• B. 3.14 x 10^-4 T
• C. 9.42 x 10^-4 T
• D. 6.28 x 10^-4 T

Answer 1

The Correct Option is B

We can use the formula for the magnetic field inside a solenoid:
\(B = \mu_0 \cdot n \cdot I\)
Given:
Length of the solenoid (L) = 50 cm = 0.5 m
Number of turns (N) = 100
Current (I) = 2.5 A
First, let's calculate the number of turns per unit length (n):
\(n = \frac{N}{L} = \frac{100}{0.5} = 200 \text{ turns/m}\)
Now, let's calculate the magnetic field at one end of the solenoid using the formula:
\(B = \mu_0 \cdot n \cdot I\)
Substituting the given values:
\(B = (4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}) \times (200 \, \text{turns/m}) \times (2.5 \, \text{A})\)
\(B = 4\pi \times 10^{-7} \times 200 \times 2.5 \, \text{T}\)
\(B = 4\pi \times 10^{-7} \times 500 \, \text{T}\)
\(B = 2\pi \times 10^{-4} \, \text{T}\)
Rounding to two decimal places, the magnetic field at one end of the solenoid is approximately \(3.14 \times 10^{-4} \, \text{T}\). Therefore, the correct answer is option (B) \(3.14 \times 10^{-4} \, \text{T}\)

Answer 2

The magnetic field \(B\) at one end of a solenoid is given by:

\(B = \frac{\mu_0 n I}{2}\)

Where:

• \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T m/A}\))
• \(n\) is the number of turns per unit length
• \(I\) is the current

Given:

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

Calculate the number of turns per unit length:

\(n = \frac{N}{l} = \frac{100}{0.5} = 200 \, \text{turns/m}\)

Now, calculate the magnetic field:

\(B = \frac{4\pi \times 10^{-7} \, \text{T m/A} \times 200 \, \text{turns/m} \times 2.5 \, \text{A}}{2}\)

\(B = \frac{4 \times 3.1416 \times 10^{-7} \times 200 \times 2.5}{2} \, \text{T}\)

\(B = 2 \times 3.1416 \times 10^{-7} \times 200 \times 2.5 \, \text{T}\)

\(B = 3.1416 \times 10^{-7} \times 1000 \, \text{T}\)

\(B = 3.1416 \times 10^{-4} \, \text{T}\)

\(B \approx 3.14 \times 10^{-4} \, \text{T}\)

Therefore, the magnetic field at one end of the solenoid is approximately \(3.14 \times 10^{-4} \, \text{T}\).