Question

A flat loop of wire consisting of a single turn of cross-sectional area 7.10 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.50 T in 1.03 s. What is the resulting induced current if the loop has a resistance of 1.60

Answer 1

The resulting induced current if the loop has a resistance of 1.60 is 0.43 mA

Step-by-step explanation:

First, we will determine the induced emf. The emf is given by

`Emf = - N(\Delta\phi)/(\Delta t)`

Where
`N` is the number of turns

`\Delta\phi` is the change in magnetic flux

and
`\Delta t` is the change in time

The change in magnetic flux is given by

`\Delta\phi = \phi _(f) - \phi _(i)`

Where
`\phi _(f)` is the final magnetic flux

and
`\phi _(i)` is the initial magnetic flux

The magnetic flux is given by

`\phi = BA`

Where
`B` is the magnitude of magnetic field

and
`A` is the area

Therefore,

`\Delta\phi = \phi _(f) - \phi _(i) = B _(f)A - B _(i) A`

`\Delta\phi = (B _(f) - B _(i))A`

From the question

`B _(f) = 1.50 T`

`B _(i) = 0.500 T`

`A = 7.10 cm^(2) = 0.00071 m^(2)`

∴
`\Delta\phi = (1.5 - 0.5)(0.00071)`

`\Delta\phi = 0.00071 Wb`

From the question

`N = 1`

`\Delta t= 1.03 s`

Hence,

`Emf = - N(\Delta\phi)/(\Delta t)` becomes

`Emf = - (1)(0.00071)/(1.03)`

`Emf = 0.000689 V`

Now, to determine the induced current

From Ohm's law

`Emf = IR`

Where
`I` is the current

and
`R` is the resistance

∴
`I =(Emf)/(R)`

From the question

`R = 1.60`

Hence,

`I =(0.000689)/(1.60)`

`I = 0.00043 A`

`I = 0.43 mA`

Hence, the resulting induced current if the loop has a resistance of 1.60 is 0.43 mA.