34 1 Visualize To develop a motional emf the magnetic field needs to be perpendicular to both so let s say its direction is into the page Solve This is a straightforward use of Equation 34 3 We have v E lB 1 0 V 1 0 m 5 0 10 T 5 2 0 10 m s 4 Assess This is an unreasonable speed for a car It s unlikely you ll ever develop a volt 34 2 Model Assume the magnetic field is uniform Visualize Please refer to Figure Ex34 2 Since a motional emf was developed the field must be perpendicular cid 71 to v The positive charges experienced a magnetic force to the left By the right hand rule the field must be out cid 71 of the page so that v B is to the left Solve This is a straightforward use of Equation 34 3 We have cid 71 Assess This is reasonable Laboratory fields are typically up to a few teslas in magnitude B E vl 0 050 V 5 0 m s 0 10 m 0 10 T 34 3 Visualize The wire is pulled with a constant force in a magnetic field This results in a motional emf and produces a current in the circuit From energy conservation the mechanical power provided by the puller must appear as electrical power in the circuit Solve a Using Equation 34 6 b Using Equation 34 6 again P F v pull v 4 0 m s P F pull 4 0 W 1 0 N P 2 2 2 v l B R B RF vl pull 2 0 20 1 0 N 4 0 m s 0 10 m 2 2 2 T Assess This is reasonable field for the circumstances given 34 4 Model Assume the field changes abruptly at the boundary between the two sections Visualize Please refer to Figure Ex34 4 The directions of the fields are opposite so some flux is positive and some negative The total flux is the sum of the flux in the two regions cid 71 Solve The field is constant in each region so we will use Equation 34 10 Take A to be into the page Then it is parallel to the field in the left region so the flux is positive and it is opposite to the field in the right region so the flux is negative The total flux is cid 71 cid 71 A B A B L L 2 0 20 m 2 0 T cos L A B R R cos R 2 0 20 m 1 0 T 0 040 Wb Assess The flux is positive because the areas are equal and the stronger field is parallel to the normal of the surface 34 5 Model Consider the solenoid to be long so the field is constant inside and zero outside Visualize Please refer to Figure Ex34 5 The field of a solenoid is along the axis The flux through the loop is only nonzero inside the solenoid Since the loop completely surrounds the solenoid the total flux through the loop will be the same in both the perpendicular and tilted cases cid 71 Solve The field is constant inside the solenoid so we will use Equation 34 10 Take A direction as the field The magnetic flux is cid 71 A loop 2 0 010 m 0 20 T to be in the same 5 6 3 10 Wb cid 71 B loop sol cos cid 71 B sol cid 71 A sol 2 r B sol cid 71 When the loop is tilted the component of B surface through which the magnetic field lines cross is increased by the same factor cid 71 in the direction of A is less but the effective area of the loop 34 6 Model Assume the field is uniform inside the rectangular region and zero outside Visualize The flux measures how much of the field penetrates the chosen surface We will break the surface up because the magnetic field has different strengths over different parts of the surface of the loop Solve For convenience we choose the normal to the loop to be into the page so it is in the same direction as the magnetic field The total flux is cid 71 cid 71 B dA cid 71 cid 71 B dA cid 71 cid 71 B dA BdA Bab cos BA loop inside rectangle outside rectangle inside rectangle inside rectangle Assess Only the region with nonzero magnetic field inside the loop contributes to the flux 34 7 Model Assume the field strength is uniform over the loop Visualize Please refer to Figure Ex34 7 According to Lenz s law the induced current creates an induced field that opposes the change in flux Solve The original field is into the page within the loop and is changing strength The induced counterclockwise current produces a field out of the page within the loop that is opposing the change This implies that the original field must be increasing in strength so the flux into the loop is increasing 34 8 Model Assume that the magnet is a bar magnet with field lines pointing away from the north end Visualize As the magnets move if they create a change in the flux through the solenoid there will be an induced current and corresponding field According to Lenz s law the induced current creates an induced field that opposes the change in flux a When magnet 1 is close to the solenoid there is flux to the left through the solenoid As magnet 1 Solve moves away there is less flux to the left The induced current will oppose this change and produce an induced current and a corresponding flux to the left By the right hand rule this corresponds to a current in the wire into the page at the top of the solenoid and out of the page at the bottom of the solenoid So the current will be right to left in the resistor b When magnet 2 is close to the solenoid the diverging field lines of the bar magnet produce a flux to the left in the left half of the solenoid and a flux to the right in the right half Since the flux depends on the orientation of the loop the flux on the two halves have opposite signs and the net flux is zero Moving the magnet away changes the strength of the field and flux but the total flux is still zero Thus there is no induced current 34 9 Visualize Please refer to Figure Ex34 9 The changing current in the solenoid produces a changing flux in the loop By Lenz s law there will be an induced current and field to oppose the change in flux Solve The current shown produces a field to the right inside the solenoid So there is flux to the right through the surrounding loop As the current in the solenoid increases there is more field and more flux to the right through the loop There is an induced current in the loop that will …
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