%3Cp%3EA%20packed%20bundle%20of%20150%20long%2C%20straight%2C%20insulated%20wire
ID: 3900704 • Letter: #
Question
%3Cp%3EA%20packed%20bundle%20of%20150%20long%2C%20straight%2C%20insulated%20wires%20forms%20a%0Acylinder%20of%20radius%20R%20%3D%200.0480cm.%20If%20each%20wire%20carries%201.30%20A%2C%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3EA.%20What%20is%20the%20magnitude%20force%20per%20unit%20length%20acting%20on%20a%20wire%0Alocated%200.240%20cm%20from%20the%20center%20of%20the%20bundle%3F%20(N%2Fm)%3C%2Fp%3E%0A%3Cp%3EB.%20What%20is%20the%20direction%20of%20the%20magnetic%20force%20per%20unit%20length%0Aacting%20on%20a%20wire%20located%200.240%20cm%20from%20the%20center%20of%20the%0Abundle%3F%3C%2Fp%3E%0A%3Cp%3EC.%20Would%20a%20wire%20on%20the%20outer%20edge%20of%20the%20bundle%20experience%20a%0Aforce%20greater%20or%20smaller%20than%20the%20value%20calculated%20in%20parts%20(a)%20and%0A(b)%3F%3C%2Fp%3E%0A%3Cp%3ED.%20Give%20a%20qualitative%20argument%20for%20your%20answer.%3Cbr%20%2F%3E%3C%2Fp%3E%0AExplanation / Answer
Use ampere's law to compute the strength of the field at a distance d=0.240 cm.
Integral(B dot dr) = u0*Ienc
By symmetry, we know B has a constant magnitude around circular loop of radius d
so B*2pi*d = u0*Ienc
B=u0*Ienc/(2pi*d) equation 1
Ienc= current enclosed by the loop,
I = the current of a single wire
Ienc= current density times area inside the loop where the current passes
Ienc=[150*I/(pi*R^2)]*pi*d^2 equation 2
F = force= q*v cross B=I*s cross B, where s = length
F=I*s*B since the length or current is perpendicular to B
F/s=I*B is the force per unit length, use equation 1 and 2 to compute
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.