Solenoid 30cm long and 8cm in diameter

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To calculate the magnetic field strength at the center of a solenoid that is 30cm long and 8cm in diameter, wound with #18 gauge copper wire, the formula B=(mu_0*N*I)/L is used. The user initially struggled to determine the current (I) from the 12-volt car battery, needing to relate voltage to resistance. It was clarified that resistance can be calculated per unit length of the wire. The user later found a new formula for current, leading to a calculated magnetic field strength of approximately 3740T. This discussion emphasizes the importance of understanding resistance and current in solenoid applications.
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You have a 12 volt car battery and a supply of #18 gauge copper wire. If you make a solenoid 30cm long and 8cm in diameter, winding two layers of wire along the length of the solenoid, what will be the strength of the magnetic field at teh center of teh solenoid when connected to the car battery? #18 gauge wire has a diameter of 1.02mm.

B=(mu_0*N*I)/L i have B=((4pi*10^7Tm/A)*(294.11turns)*(I))/(.3m) the resistance of copper is 1.7*10^-8 but I am not sure how to get from the 12 volts to the current (I) i have worked this problem a few different ways and i haven't come up with anything that works please help me out
 
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You can find the current just by relating the voltage to the resistance, usually the resistance of a material is given per unit length.. is that the case?
 
answer

i believe i can do that i found a new formula I=(density*Length)/area * volts and when i worked the problem it came out to be something about 3740T.
 

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