Question

The magnitude of magnetic field inside a solenoid of length 0.3 m having 800 turns carrying a current of 6 A is

• A. 2.03 T
• B. 60.3 mT
• C. 20 mT
• D. 6.03 T

Hint:

The formula \(B = \mu_0 n I\) is for an ideal, infinitely long solenoid. However, it provides a very good approximation for the field near the center of a real solenoid as long as its length is much greater than its diameter. In exams, unless specified otherwise, use this ideal formula.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A solenoid is a coil of wire that produces a nearly uniform magnetic field in its interior when a current is passed through it. The strength of this magnetic field depends on the number of turns per unit length, the current, and the permeability of the medium inside.

Step 2: Key Formula or Approach:
The magnitude of the magnetic field \(B\) inside a long solenoid is given by the formula: \[ B = \mu_0 n I \] where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)), \(I\) is the current, and \(n\) is the number of turns per unit length. The number of turns per unit length \(n\) is calculated as \(n = \frac{N}{L}\), where \(N\) is the total number of turns and \(L\) is the length of the solenoid.

Step 3: Detailed Explanation:
Given data:
Length of the solenoid, \(L = 0.3 \, \text{m}\).
Total number of turns, \(N = 800\).
Current, \(I = 6 \, \text{A}\).
Calculation:
First, calculate the number of turns per unit length, \(n\): \[ n = \frac{N}{L} = \frac{800}{0.3} = \frac{8000}{3} \, \text{turns/m} \] Now, substitute the values into the magnetic field formula: \[ B = \mu_0 n I = (4\pi \times 10^{-7}) \times \left(\frac{8000}{3}\right) \times 6 \] \[ B = (4\pi \times 10^{-7}) \times (8000 \times 2) \] \[ B = 4\pi \times 10^{-7} \times 16000 \] \[ B = 64000 \pi \times 10^{-7} = 6.4\pi \times 10^{-3} \, \text{T} \] Now, we calculate the numerical value using \(\pi \approx 3.14159\): \[ B \approx 6.4 \times 3.14159 \times 10^{-3} \, \text{T} \] \[ B \approx 20.106 \times 10^{-3} \, \text{T} \] This is equal to 20.106 mT. The closest option is 20 mT.

Step 4: Final Answer:
The magnitude of the magnetic field inside the solenoid is approximately 20 mT.