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Problem 30 Hard Difficulty

Consider the arrangement shown in Figure P20.30 where $R=6.00 \Omega, \ell=1.20 \mathrm{m},$ and $B=2.50 \mathrm{T}$ . (a) At what constant speed should the bar be moved to produce a current of 1.00 $\mathrm{A}$ in the resistor? (b) What power is delivered to the resistor? $(\mathrm{c})$ What magnetic force is exerted on the moving bar? (d) What instantaneous power is delivered by the force $F_{\mathrm{app}}$ on the moving bar?

Answer

a. \mathrm{v}=2 \mathrm{m} / \mathrm{s}
b. P=6 W
c. \mathrm{F}=3 \mathrm{N}
d. P=6 W

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Video Transcript

for part A were asked to find the constant speed that the bar should be moved at to produce it current I of one ampere. So to do this we can use the fact that the potential is equal to the magnetic field which we know times the length, which we also know multiplied by the velocity which were asked to find. But potential is also equal to kind well played by resistance from home slop. Therefore, the velocity is equal to the magnetic field times the length. Excuse me. Sorry had that inverted? The velocity is equal to the current times the resistance divided by the magnetic field times the length plugging those values into this equation, we find that the velocity is equal to two meters per second, so we can go ahead and box two meters per second in as our solution for part a starting a new page for part B for part B, it says what power is delivered to the resistance? Er willpower is equal to the current times, the potential which is equal to the current squared times the resistance clicking those values into this equation. We find that this is equal to six in the units for power. What's so you can box set in his solution for B part C asked us to find the force that is exerted on the moving bar. So force here from magnetic field is equal to the current, multiplied by the length of the bar cross the magnetic field. So this is the current times The length of the bar times magnetic field and cross product is the sine of the angle between them. That's 90 degrees. So this is equal to current times the length times, the magnetic field plugging those values into this expression, we find that the force here is equal to three Nunes. This weekend box I did was our solution for sea and then for the very last part, were asked to find the instantaneous power delivered by the force applied on the moving bar. Well, instantaneous power is equal to the force being applied, which is the force from the bar magnet. Oh, multiplied by the velocity. And let's use the, uh, same notation we used for part a here because this is the velocity we found in part A. We found this to be two meters per second So the force applied, which we found in part, uh, part C to be three Newton's multiplied by two meters per second. Give us gives us a value of six. Once Macon box. That is our solution for party.

University of Kansas
Top Physics 102 Electricity and Magnetism Educators
Elyse G.

Cornell University

LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Jared E.

University of Winnipeg

Physics 102 Electricity and Magnetism Bootcamp

Lectures

Join Bootcamp