Calculating Solenoid Turns: Magnetic Field, Current, and Dimensions

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To calculate the number of turns on a solenoid with a diameter of 20 cm, a length of 248 cm, and a current of 4.9 A producing a magnetic field of 3.3 × 10^-3 T, the formula B = μ₀nI can be used, where n is the number of turns per unit length. The diameter is not a factor in this calculation, and the focus should be on the relationship between the magnetic field, current, and turns per unit length. To find n, rearrange the formula to n = B / (μ₀I). The discussion emphasizes the importance of understanding the magnetic field's dependence on current and turns rather than the solenoid's dimensions.
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(a) How many turns of wire would be on a solenoid carrying a current 4.9 A if the solenoid is 20 cm in diameter, 248 cm long, and the field at the center is 3.3 × 10-3 T?



The magnetic field of a solenoid is the permeability of free space times the number of turns per unit length times the current.



Now How do i find the Number of turns per unit length, Is it Length/Diameter * The field of Center? Thanks
and i know n = N/L so how do i find that
 
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Okay so Basically i tried to use N = 2RB/ (uoI) to get the answer but that does not work
 
Alt+F4 said:
(a) How many turns of wire would be on a solenoid carrying a current 4.9 A if the solenoid is 20 cm in diameter, 248 cm long, and the field at the center is 3.3 × 10-3 T?



The magnetic field of a solenoid is the permeability of free space times the number of turns per unit length times the current.



Now How do i find the Number of turns per unit length, Is it Length/Diameter * The field of Center? Thanks
and i know n = N/L so how do i find that
The diameter of a long solenoid does not matter. The magnetic field (ideally) is B=(uo)nI . Solve for n .
 
Chi Meson said:
The diameter of a long solenoid does not matter. The magnetic field (ideally) is B=(uo)nI . Solve for n .
thanks alot
 
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