A solenoid with 200 turns has a length of 0.5 m and carries a current of 2 A. What is the magnetic field inside the solenoid? (2021)
Practice Questions
1 question
Q1
A solenoid with 200 turns has a length of 0.5 m and carries a current of 2 A. What is the magnetic field inside the solenoid? (2021)
0.4 T
0.8 T
1.0 T
1.6 T
Magnetic field B = μ₀ * (N/L) * I = (4π x 10^-7 Tm/A) * (200/0.5) * 2 = 0.8 T.
Questions & Step-by-step Solutions
1 item
Q
Q: A solenoid with 200 turns has a length of 0.5 m and carries a current of 2 A. What is the magnetic field inside the solenoid? (2021)
Solution: Magnetic field B = μ₀ * (N/L) * I = (4π x 10^-7 Tm/A) * (200/0.5) * 2 = 0.8 T.
Steps: 8
Step 1: Identify the formula for the magnetic field inside a solenoid, which is B = μ₀ * (N/L) * I.
Step 2: Understand the variables in the formula: μ₀ is the permeability of free space (4π x 10^-7 Tm/A), N is the number of turns (200), L is the length of the solenoid (0.5 m), and I is the current (2 A).
Step 3: Substitute the values into the formula: B = (4π x 10^-7 Tm/A) * (200/0.5) * 2.
Step 4: Calculate N/L: 200 turns divided by 0.5 m equals 400 turns/m.
Step 5: Now substitute this value back into the equation: B = (4π x 10^-7 Tm/A) * 400 * 2.
Step 6: Calculate the product: 400 * 2 = 800.
Step 7: Now multiply: B = (4π x 10^-7 Tm/A) * 800.
