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) \(3.14 \times 10^{-4} \, \text{T}\)