126k views
2 votes
A 51.0 cm-long solenoid 1.35 cm in diameter is to produce a field of 0.450 mt at its center. how much current should the solenoid carry if it has 715 turns of wire?

1 Answer

5 votes

Final answer:

To calculate the current a solenoid should carry to produce a specified magnetic field, use the formula B = μ0 * n * I, and solve for I with the given solenoid dimensions and desired magnetic field strength.

Step-by-step explanation:

The student is asking how much current should a solenoid carry if it is to produce a magnetic field at its center. The solenoid in question is 51.0 cm in length, 1.35 cm in diameter, has 715 turns, and is required to produce a magnetic field of 0.450 mT (millitesla) at its center.

The magnetic field B inside a solenoid is given by the formula B = μ0 * n * I, where μ0 is the permeability of free space (4π * 10-7 Tm/A), n is the number of turns per meter, and I is the current.

To find the current I, we can rearrange this formula to I = B / (μ0 * n). The number of turns per meter n can be calculated by dividing 715 turns by the length of the solenoid in meters (0.51 m), giving us n = 715 / 0.51 turns/m. Then, we plug in the values for B (0.450 mT or 0.450 * 10-3 T) and μ0 to find I.

After calculating, we obtain the required current the solenoid should carry to produce the desired magnetic field.

User Tkrishtop
by
9.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.