Hello. In the solution in problem 6 out of Sherway 8th ed. ch20 (Problem: A 400-

Question

Hello. In the solution in problem 6 out of Sherway 8th ed. ch20 (Problem: A 400-turn solenoid of length 36.0 cm and radius of 3.00 cmcarries a current of 5.00 A. Find the magnetic field strengthinside the coil at its midoint and the magnetic flux through acircular cross sectional area of the solenoid at itsmidpoint)
I got the same answer for the magnetic field... B = 6.98mT
In for the flux, the equation given is flux= B A the solution gives this, then proceeds to divide B by A to gettheir answer, while I multiplied them and obviously got a differentanswer of 1.97 x 10-5Wb I was wondering why they divided B by A?? In the solution in problem 6 out of Sherway 8th ed. ch20 (Problem: A 400-turn solenoid of length 36.0 cm and radius of 3.00 cmcarries a current of 5.00 A. Find the magnetic field strengthinside the coil at its midoint and the magnetic flux through acircular cross sectional area of the solenoid at itsmidpoint)
I got the same answer for the magnetic field... B = 6.98mT
In for the flux, the equation given is flux= B A the solution gives this, then proceeds to divide B by A to gettheir answer, while I multiplied them and obviously got a differentanswer of 1.97 x 10-5Wb I was wondering why they divided B by A??

Explanation / Answer

   first we find the magnetic field generatedby the current in the solenoid as    B = o n I        = ........... T    then flux through each turn on the solenoidwill be    B = B A cos          = B Acos0o          = ...........T . m2

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