Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.60 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.40 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.80

Answer 1

The induced current is
`I = 5.72*10^(-4 ) \ A`

Step-by-step explanation:

From the question we are told that

The cross-sectional area is
`A = 8.60 \ cm^2 = (8.60 )/(10000) = 8.60 *10^(-4) \ m`

The initial value of magnetic field is
`B_1 = 0.500 \ T`

The value of magnetic field at time t is
`B_f = 2.40 \ T`

The number of turns is N = 1

The time taken is
`dt`= 1.02 \ s

The resistance of the loop is
`R = 2.80\ \Omega`

Generally the induced emf is mathematically represented as

`e = - (d \phi)/(dt )`

Where
`d \phi` is the change n the magnetic flux which is mathematically represented as

`d \phi = N *A * d B`

Where
`dB` is the change in magnetic field which is mathematically represented as

`d B = B_f - B_i`

substituting values

`d B = 2.40 - 0.500`

`d B = 1.9 \ T`

So

`d \phi = 1 * 1.9 * 8.60 *10^(-4)`

`d \phi = 1.63*10^(-3) \ T`

So

`e = - (1.63 *10^(-3))/( 1.02 )`

`e = - 1.60*10^(-3) \ V`

Here the negative only indicates that the emf is acting in opposite direction of the motion producing it so the magnitude of the emf is

`e = 1.60*10^(-3) \ V`

Now the induced current is evaluated as follows

`I = (e)/(R )`

substituting values

`I = (1.60 *10^(-3))/(2.80 )`

`I = 5.72*10^(-4 ) \ A`