An emf of 25.0 mV is induced in a496 turn coil when the current is changingat a

Question

An emf of 25.0 mV is induced in a496 turn coil when the current is changingat a rate of 10.0 A/s. What is themagnetic flux through each turn of the coil at an instant when thecurrent is 6.00 A?
T·m2 Please show all steps with the values given inthe problem. Also, please show solutions completely toreceive a rating of lifesaver. An emf of 25.0 mV is induced in a496 turn coil when the current is changingat a rate of 10.0 A/s. What is themagnetic flux through each turn of the coil at an instant when thecurrent is 6.00 A?
T·m2 Please show all steps with the values given inthe problem. Also, please show solutions completely toreceive a rating of lifesaver. Please show all steps with the values given inthe problem. Also, please show solutions completely toreceive a rating of lifesaver.

Explanation / Answer

                       Giventhat the number of turns in the coil is N = 496           Theinduced emf is = 25.0 V           Therate of chage of current is dI/dt = 10.0 A/s           Thecurrent in the each turn is I = 6.00 A      ---------------------------------------------------------------------------    The emf of the coil is = d(B.A)/dt                                     butB = 0N*I/ 2r                    where r is the radius of the coil                 then = ( 0N/ 2r)*r2*dI/dt                            = ( 0N/ 2 )*r*dI/dt                 25.0V = (4*10-7 T.m/A)(496)*3.14*r*(10.0A/s)/2            Then we get r = ------- m      The magnetic flux through each turnof the coil = B.A                                                                               = ( 0*I/ 2r )*(r2)                                                                               = ( 0*I/ 2 )*(r)                                                                               = ----------- Wb                                                                                                                                Giventhat the number of turns in the coil is N = 496           Theinduced emf is = 25.0 V           Therate of chage of current is dI/dt = 10.0 A/s           Thecurrent in the each turn is I = 6.00 A      ---------------------------------------------------------------------------    The emf of the coil is = d(B.A)/dt                                     butB = 0N*I/ 2r                    where r is the radius of the coil                 then = ( 0N/ 2r)*r2*dI/dt                            = ( 0N/ 2 )*r*dI/dt                 25.0V = (4*10-7 T.m/A)(496)*3.14*r*(10.0A/s)/2            Then we get r = ------- m      The magnetic flux through each turnof the coil = B.A                                                                               = ( 0*I/ 2r )*(r2)                                                                               = ( 0*I/ 2 )*(r)                                                                               = ----------- Wb                                                                                                                     

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