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Home Q & A Board Science-Related Homework Help Physics
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ppk ppk
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12 years ago
How many turns should a 10-cm long solenoid have if it is to generate a 1.5 × 10-3 T magnetic field on 1.0 A of current?
A) 12
B) 15
C) 119
D) 1194
E) 3183
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#1
12 years ago
First the equation for the magnetic field oven off by a solenoid is B=\({\mu}[sub]0[/sub]\)\(\)(N/L)I where B is the magnetic field \({\mu}[math=inline][sub]0[/sub]\)[/math] is the magnetic constant equal to approximately 4\({\pi}[math=inline]\)[/math]x10-7 V*s/(A*m) N is number of turns and L is length in meters and I is the number of amps.  So if we plug in the numbers we get 1.5x10^-3=(4\({\pi}[math=inline]\)[/math]x10^-7)(N/.1)*1 so with some algebra we find N=((1.5*10^-3)(.1))/(4\({\pi}\)\(\)*10^-7) so N=119 or C.)
goober123
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