Each of two parallel wires separated by 6.00 mm carries a 40-A current. These tw

Question

Each of two parallel wires separated by 6.00 mm carries a 40-A current. These two currents are in the same direction. Determine the magnitude of the magnetic field at a point that is 5.0 mm from each of the wires. [Think three dimensionally.] 2.6 mT 3.8 mT 1.9 mT 1.4 mT 3.2 mT A certain rod has a length of 96.0 cm and moves southward through a uniform magnetic field that is directed downward. The ends of the rod are oriented to the east and west and the magnitude of the field is 0.0480 T. If the induced emf between the ends of the rod is 38.0 mV, what is the speed of the rod? 17.5 cm/s 24.3 cm/s 49.8 cm/s 67.4 cm/s 82.5 cm/s

Explanation / Answer

20)

B1 = uI/(2*pi*r) = 1.26*10^-6*40/(2*pi*0.005) = 1.6*10^-3 <-------magnitude

B2 =  1.26*10^-6*40/(2*pi*0.005) = 1.6*10^-3 <----magnitude

So, Bnet = B1 + B2 <------vector addition

So, Bnet = 2*1.6*10^-3*cos(asin(3/5)) = 2.6*10^-3 T = 2.6 mT <---------answer

21)

Induced emf = BIv

where B = magnetic field= 0.048 T

I = length = 0.96 m

So, 38*10^-3 = 0.048*0.96*v

So, v = 0.825 m/s = 82.5 cm/s <----------answer

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.