Faraday CT'sF-1 In which situation is the magnetic flux through the loop the smallest?Answer: When the loop is face-on to the field, the flux is maximum. When the loop is edge-on to the field, the flux through the loop is zero. So the flux is minimum in case (A) when the loop is edge-on to the field. Magnetic flux is B A^F =.F-2 A loop of wire is moving rapidly through a uniform magnetic field as shown. True (A) or False(B): there is a non-zero emf induced in the loop. Answer: False. The magnetic flux through the loop is not changing (constant B-field, constant area A, constant angle between B-field and plane of loop), so the emf is zero.F-3 A loop of wire is spinning rapidly about a stationary axis in uniform magnetic field as shown. constant, uniformB-field(A)(B)(C)area Aedge-onarea Aface-onarea Atilted 60°(D) Same flux in all three situationsTrue(A) or False(B): there is a non-zero emf induced in the loop.Answer: True. There is a changing flux through the loop, since the angle between the B-field and the plane of the loop is changing. A changing flux creates an emf.F-4 The magnetic flux through a loop of wire is shown. At which point is the emf induced in theloop a maximum?Answer: The rate of change of flux (and hence the emf) is maximum at point B. F-5 A loop of wire is sitting in a uniform, constant magnet field as shown. Suddenly, the loop isbent into a smaller area loop. During the bending of the loop, the induced current in the loop is A) zero B) clockwise C) counterclockwisetimeflux ABDCAnswer: The flux is decreasing as the loop area decreases. To fight the decrease, we want the induced B to add to the original B. By the modified right hand rule, a clockwise induced current will make an induced B into the page, adding to the original B. The answer is: Clockwise.F-6 A bar magnet is positioned below a horizontal loop of wire with its North pole pointing toward the loop. Then the magnet is pulled down, away from the loop. As viewed from above, is the induced current in the loop clockwise or counterclockwise?B(in)B(in)A) clockwise B) counter-clockwiseNSeyeballAnswer: The B-field from a bar magnet points out of the North pole. As seen from above, the field through the loop is out (toward the observer). As the magnet is pulled away, the flux is decreasing. To fight the decrease, the induced B-field should add to the original B-field, and also be out (toward the observer). The induced current will be (B), counterclockwise, in order tomake an induced B-field out.F-7 A square loop is rotating in a fixed, external magnetic field into the page. At the instant shown, the loop is out of the plane of the page with left side of the loop above the page and coming out of the page, the right side is below the page going away. The direction of the induced current is ...Answer: At the moment shown, the flux through the loop is decreasing (since the amount of B-field "threading" the loop is decreasing.) To fight the decrease, the induced field should add to the original field. The answer is B, Clockwise.F-8 A loop of wire is near a long straight wire which is carrying a large current I, which is decreasing. The loop and the straight wire are in the same plane and are positioned as shown. The current induced in the loop is A) counter-clockwise B) clockwise C) zero.axis of rotationB(in)A BC) NeitherAnswer: CW. At the position of the loop, the B-field created by the long wire is into the page. The flux is decreasing, so Lenz's law says the induced B-field should into the page, to add to the original flux and fight the decrease. Modified right-hand-rule says that to make B point into page, we need a clockwise induced current.F-9 Two loop of wires labeled A and B are placed near each other as shown. A large current I inloop A is suddenly turned on. This causes an induced current in loop B which causes..A) a net repulsive force - the two loops repel.B) a net attractive force - the two loops attract.C) whether the force is attractive or repulsive depends on whether the current in first loop is CW or CCW.D) no net force.Answer: A) a repulsive force, the two loops repel.Assume a CCW current I in Loop A. When the current I is turned on, there is an increasing B-field pointing out of the page in the middle of both loops A and B. This increasing B-field outward induces a CW current in loop B so as to create an induced B-field into the page (fightingthe increase in flux). Now, you have to remember that anti-parallel currents repel; parallel currents attract. The two currents are anti-parallel, so the loops repel.If the current in loop A was the other way (CW), then everything would be reversed and the two currents would still be anti-parallel.I to the right, but decreasing.loopF-10 A wire loop, moving right, enters a region where there is a constant, uniform magnetic field pointing into the page.As the loop enters the B-field, the current induced in the loop is A) CW B) CCWAs the loop enters the B-field, the direction of the net force on the loop is ... A) right B) left C) up D) downE) there is no net forceAnswers: Question 1: A) CCW. As the loop enters the field, the flux increases. To fight the increasing flux,the induced B-field (out of the page) should be opposite the original B (into the page). A CCW current will create an induced B out of the page. This current is an example of an "eddy current".Question 2: B) left.Method I (the easy way ): The magnetic forces on eddy currents are always in the direction which opposes the motion. (Otherwise you could make a perpetual motion machine).Method II (harder way): Work out all the directions. As the loop enters the field, the flux thru the loop is increasing. By Lenz's law, the induced current in the loop will be CCW [so that the induced B-field (out of the page) fights the change in flux]. There will be a magnetic force on the right side of the loop . The direction of that force is to the left.BBF-11 The vertical wire shown is moving to the right in a uniform magnetic field which is into the page. There is a current upward, meaning that there is a flow of electrons downward. The fixed positive ions in the wire are moving to the right, along with the wire. The negative conduction electrons are moving to the right and down. What is the direction of the net magnetic force on this segment of the wire?A) Up B) Down C) Left D) Right E) None of theseAnswer: C) The net force on the
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