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a flat loop of wire consisting of a single turn of cross-sectional area 7.50 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 t to 1.70 t in 1.00 s. what is the resulting induced current if the loop has a resistance of 2.90 ?

Respuesta :

The resulting induced current if the loop has a resistance of 2.90 is 0.80  mA.

A magnetic field is a vector subject that describes the magnetic having an impact on transferring electric fees, electric currents, and magnetic substances. A shifting charge in a magnetic field experiences a pressure perpendicular to its very own velocity and to the magnetic field.

Calculation:-

Induced EMF    E  =  - dφ/dt  =  - N * A * dB/dt

here no. of turns  N = 1

change in magnetic field  dB  =  B2 - B1

                         =  2.50 T - 0.500 T

                                        = 2.00 T

Corresponding time    dt  =  1.00 s

Area of loop   A = 8.00 cm2   =  8.00 * 10-4  m2

E = - 1 * 8.00 * 10-4 * 2.00 / 1.00

=  - 1.60 * 10-3     V

=  - 1.60  mV

- ve sign can be discarded as it indicates direction only.next step

b.  Current I  =  E / R

 =  1.60 mV / 2.00 Ω

     = 0.80  mA

Learn more about the magnetic fields here:-https://brainly.com/question/14411049

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