Question

A solenoid having 60 turns and length 15 cm produces magnetic field of 2.4 × 10^–3 T, Find the current in the solenoid

Answer 1

Correct Answer: 100

Given:

• Number of turns (\( N \)) = 60
• Length of the solenoid (\( l \)) = 15 cm = 0.15 m
• Magnetic field (\( B \)) = \( 2.4 \times 10^{-3} \, \text{T} \)
• Permeability of free space (\( \mu_0 \)) = \( 4\pi \times 10^{-7} \, \text{T·m/A} \)

Step 1: Formula for Magnetic Field in a Solenoid

The magnetic field inside a solenoid is given by the formula:

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

Where:

• \( B \): Magnetic field
• \( \mu_0 \): Permeability of free space
• \( n = \frac{N}{l} \): Number of turns per unit length
• \( I \): Current in the solenoid

Step 2: Substituting Values

First, calculate the number of turns per unit length: \[ n = \frac{N}{l} = \frac{60}{0.15} = 400 \, \text{turns/m} \] Now substitute into the formula: \[ B = \mu_0 \cdot n \cdot I \] \[ 2.4 \times 10^{-3} = (4\pi \times 10^{-7}) \cdot 400 \cdot I \]

Step 3: Solve for Current (\( I \))

Rearranging the formula for \( I \): \[ I = \frac{B}{\mu_0 \cdot n} \] Substituting the values: \[ I = \frac{2.4 \times 10^{-3}}{(4\pi \times 10^{-7}) \cdot 400} \] Simplify: \[ I = \frac{2.4 \times 10^{-3}}{1.6 \times 10^{-4}} \] \[ I = 100 \, \text{A} \]

Step 4: Final Answer

The current in the solenoid is \( \boxed{100 \, \text{A}} \).