A 25.0 ohm resistor is sliding without friction down an inverted U-shaped conduc

Question

A 25.0 ohm resistor is sliding without friction down an inverted U-shaped conducting frame. The frame has negligible resistance. The resistor and frame are immersed in a uniform magnetic field of 45.0 T directed perpendicular to the frame. See Figure 1, where the magnetic field is directed out of the diagram. The distance b is 0.250 m and the resistor slides down with constant speed v. The mass of the sliding resistor (along with its connecting wires) is 12.5 g. a. Give the formula for the magnetic flux Phi enclosed by the circuit ii terms of b, h and B. Express the rate of change of magnetic flux through this. circuit in terms of v, then use Faraday's law to express the emf induced in the circuit in terms of y. (Take the counterclockwise direction around the circuit as positive.) b. Give a formula for the current i induced in the circuit in terms of y then find the value of v by requiring that the upward magnetic force on the falling resistor just balances its weight (Treat the falling resistor as if it were a straight wire.) What is the direction of the current around the circuit?

Explanation / Answer

a) Suppose at any time instance t the given configuration is occured. Then,

There are two forces working on the resistor, force due to gravity and force due to magnetic field.

The flux enclosed by the circuit will be,

Phi = mangetic field times Area = B*bh

rate of change of flux = d_phi/dt = B*b* dh/dt = B*b*v

the emf = - rate of change of flux = - vBb

b)

Current induced = emf/resistance = - vBb/R

So the magnetic force = i*l*B = -(v*B*b/R)*b*B

Gravity force = mg

Hence , mg = (v*B*b/R)*b*B

v = mgR/(bB)^2 = 12.5*10^-3*9.81*25/(0.25*45)^2 = 0.024 m/s

The direction of the current is clockwise.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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