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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?

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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.

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