A medical physicist is designing a new imaging device that requires a strong, uniform magnetic field. They are considering using a solenoid with 500 turns and a length of 0.25 m. If a current of 2.0 A is passed through the solenoid, what is the magnitude of the magnetic field inside the solenoid? (Permeability of free space, μ₀ = 4π x 10⁻⁷ T·m/A)
- A1.0 x 10⁻³ T
- B8.0 x 10⁻³ T
- C2.5 x 10⁻³ T
- D5.0 x 10⁻³ T
Show answer & explanationAnswer & explanation
Correct answer: D. 5.0 x 10⁻³ T
The magnetic field inside a solenoid is given by the formula B = μ₀ * (N/L) * I, where μ₀ is the permeability of free space, N is the number of turns, L is the length of the solenoid, and I is the current. Plugging in the values: B = (4π x 10⁻⁷ T·m/A) * (500 turns / 0.25 m) * 2.0 A. B = (4π x 10⁻⁷) * (2000) * 2.0 = 16000π x 10⁻⁷ T = 1.6π x 10⁻³ T. Since π ≈ 3.14, B ≈ 1.6 * 3.14 x 10⁻³ T ≈ 5.024 x 10⁻³ T. This is approximately 5.0 x 10⁻³ T.
Why the other options are wrong
- A. This value is too low, perhaps due to an arithmetic error or miscalculation of N/L.
- B. This value is too high, indicating an error in calculation or formula application.
- C. This value is incorrect, possibly from using an incorrect constant or calculation.
Magnetic Field of a Solenoid
A solenoid is a coil of wire acting as a magnet when current passes through it. The magnetic field inside a long solenoid is approximately uniform and its magnitude is directly proportional to the number of turns per unit length and the current.
- Formula: B = μ₀ * (N/L) * I.
- μ₀ is permeability of free space (constant).
- N is number of turns, L is length, I is current.
- Field is uniform and strong inside, weak outside.
Memory trick: B equals 'mu not' times 'N over L' times 'I'.