A 27.0 cm long solenoid with 940 turns and 1.25 cm in diameter
Magnetic Field of Solenoid Calculation
A 27.0 −cm long solenoid 1.25 cm in diameter is to produce a field of 0.390 T at its center. How much current should the solenoid carry if it has 940 turns of the wire?
Final answer: The current needed to create a 0.390 T field in a solenoid with 940 turns and 0.27m in length at its center is approximately 1.765 Amps.
Explanation: The magnetic field at the center of a solenoid can be calculated using the formula B = μ0*n*I, where B is the magnetic field strength, μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current. Given the values of B, n (940 turns over 0.27m of length), and μ0 (4π*10^-7 T m/A), we can solve for I by rearranging the formula to I = B / (μ0*n). Inserting the given values into our formula, we find that the current needed to create a 0.390 T field in this solenoid at its center is approximately 1.765 Amps.
How is the current needed to create a magnetic field in a solenoid calculated? The current needed to create a magnetic field in a solenoid is calculated using the formula B = μ0*n*I, where B is the magnetic field strength, μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.