Problems 9 and 10 refer to the same diagram: A loop of wire with length 0.75 cm

Question

Problems 9 and 10 refer to the same diagram: A loop of wire with length 0.75 cm and width 0.35 cm is removed from a region of constant magnetic field 0.38 T. The loop of wire has resistance of 1 ohm total. The wire is to be removed with a velocity of 3.5 m/s. What is the induced current, and in which direction does it flow? How large of a force must be exerted? The same loop of wire were to be inserted into the magnetic field at the same velocity. What is the induced current, and in which direction does it flow? How large of a force must be exerted?

Explanation / Answer

Induced emf

= d(BA)/dt

=0.39*dA/dt

=0.39*1.350*dx/dt

=0.39*1.350*3.5

=1.843 V

Induced current,

I = EMF / R = 1.843 A.

Force exerter,

F = q (vxB)

= I (LxB)

=ILB sin(theta)

=1.843*1.350*0.38*1

=0.945 N.

10.Here

F = 0, I = 0. Since there's no change of magnetic field.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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