A magnetic field is perpendicular to the plane of asingle-turn circular coil. Th

Question

A magnetic field is perpendicular to the plane of asingle-turn circular coil. The magnitude of the field is changing,so that an emf of 0.61 V and a current of2.9 A are induced in the coil. The wire isthe re-formed into a single-turn square coil, which is used in thesame magnetic field (again perpendicular to the plane of the coiland with a magnitude changing at the same rate). What emf andcurrent are induced in the square coil? emf ______ V current____A A magnetic field is perpendicular to the plane of asingle-turn circular coil. The magnitude of the field is changing,so that an emf of 0.61 V and a current of2.9 A are induced in the coil. The wire isthe re-formed into a single-turn square coil, which is used in thesame magnetic field (again perpendicular to the plane of the coiland with a magnitude changing at the same rate). What emf andcurrent are induced in the square coil? emf ______ V current____A

Explanation / Answer

   according to Faraday’s law the emf inducedin either single-turn coil is given by
    = - N ( / t)    as N = 1     = - ( /t)    the flux is given by     = B A cos        = B Acos0o        = B A    were the angle between the field and thenormal to the plane of the coil is =0o    because the field is perpendicular to theplane of the coil and, hence, parallel to the normal.    With this expression for the flux,Faraday’s law becomes     = - ( /t)       = - ((B A) /t)       = - A (B / t)………… (1)    in this expression we have recognized thatthe area A does not change in time and have    eparated it from the magnitude B of themagnetic field.    we will apply this form of Faraday’slaw to each coil.    the current induced in either coil dependson the resistance R of the coil, as well as the emf.    according to Ohm’s law, the current Iinduced in either coil is given by    i = /R ………..(2)    applying Faraday’s law in the form ofequation 1 to both coils, we have    square = - (Asquare)(B / t)                = - L2 (B / t)    circle = -(Acircle) (B / t)                = - r2 (B / t)    area of a square of side L is L2,and the area of a circle of radius r is r2    the rate of change B / t ofthe field magnitude is the same for both coils    dividing these two expressionsgives    square/ circle = L2 / r2 ………… (3)    as the same wire is used for both coils, sowe know that the perimeter of the square equals the    circumference of the circle so we canwrite    4 L = 2 r    L / r = 2 / 4    substituting this result into equation (3)gives    square/ circle = / 4    square = ( / 4)circle               = ( / 4) (0.61 V)                =......... V………… (4)    using ohm’s law as given in equation(2), we find for the induced currents that    Isquare / Icircle= (square / R)/ (circle / R)                           =square / circle    here have used the fact that the same wirehas been used for each coil, so that the resistance R is    the same for each coil    using the result in equation (4) gives    Isquare / Icircle= square / circle                          = / 4    Isquare = ( / 4)Icircle               = ( / 4) (2.9 A)                =........ A    the rate of change B / t ofthe field magnitude is the same for both coils    dividing these two expressionsgives    square/ circle = L2 / r2 ………… (3)    as the same wire is used for both coils, sowe know that the perimeter of the square equals the    circumference of the circle so we canwrite    4 L = 2 r    L / r = 2 / 4    substituting this result into equation (3)gives    square/ circle = / 4    square = ( / 4)circle               = ( / 4) (0.61 V)                =......... V………… (4)    using ohm’s law as given in equation(2), we find for the induced currents that    Isquare / Icircle= (square / R)/ (circle / R)                           =square / circle    here have used the fact that the same wirehas been used for each coil, so that the resistance R is    the same for each coil    using the result in equation (4) gives    Isquare / Icircle= square / circle                          = / 4    Isquare = ( / 4)Icircle               = ( / 4) (2.9 A)                =........ A                          = / 4    Isquare = ( / 4)Icircle               = ( / 4) (2.9 A)                =........ A

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