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 December 9th, 2009, 06:38 AM #1 Newbie   Joined: Dec 2009 Posts: 4 Thanks: 0 Induced emf problem A large cylindrical loop of 288 turns and radius 67.0 cm carries a current of 70.0 A. A small square loop of 36 turns and 1.00 cm on a side is placed at the center of the large loop. If the current in the large loop drops to 0 in 0.0470 s, find the induced emf in the small loop. (Assume that the magnetic field in the region of the square loop is uniform.)
 June 8th, 2010, 09:51 AM #2 Senior Member   Joined: Jan 2009 Posts: 345 Thanks: 3 Re: Induced emf problem N(magnetic flux) = N(BA) where B is the magnetic strength, N is the number of turns and A is the Area B (magnetic strength) = (µI)/2?R where µ is the permitivity of free space, I is the current and R is the radius from the center of the wire. A=Qt where Q is the charge and t is the time Can you move on from here. Also, by lenz's law, the current induced in a wire tends to oppose normal direction when flux changes or the magnetic strength changes
 June 10th, 2010, 05:18 AM #3 Senior Member   Joined: Jan 2009 Posts: 345 Thanks: 3 Re: Induced emf problem $\text V(voltage)= Blv, where B is the magnetic strength, l is the length of the$ $\text wire or loop between 2 points and v is the velocity.$ $\text Recall that F= Bqv$ $\text And E= \frac{F}{q} so Eq = F where E is the electric field, F is the force and q is the$ $\text charge.$ $\text and that V or electric potential= work done per charge = \frac{F*d}{q} = Ed$ $\text so V(voltage)= \frac{F*d}{q}$ Lets take a step back $\text F= Bqv$ $\text F*d= (Bqv)*d (multiply both sides by d)$ $\text \frac{F*d}{q}= \frac{Bqv*d}{q} (Divide both sides by the charge q)$ Can you move on from here? $\text V(voltage)= \frac{F*d}{q} = Bvd where d is the distance from one point of the wire to another$ $\text point of the wire, a length}$ $\text V(voltage)= \frac{F*d}{q} = Bvl$

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