Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 G to 1.60 M in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20 Ω?

Answer 1

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of
`1.20 \ \Omega`

Answer:

The current is
`I = 0.0007 41 \ A`

Step-by-step explanation:

From the question we are told that

The area is
`A = 8.00 \ cm^2 = 8.0 *10^(-4) \ m^2`

The initial magnetic field at
`t_o = 0 \ seconds` is
`B_i = 0.500 \ T`

The magnetic field at
`t_1 = 0.99 \ seconds` is
`B_f = 1.60 \ T`

The resistance is
`R = 1.20 \ \Omega`

Generally the induced emf is mathematically represented as

`\epsilon = A * (B_f - B_i )/( t_f - t_o )`

=>
`\epsilon = 8.0 *10^(-4) * (1.60 - 0.500 )/( 0.99- 0 )`

=>
`\epsilon = 0.000889 \ V`

Generally the current induced is mathematically represented as

`I = (\epsilon)/(R )`

=>
`I = (0.000889)/( 1.20 )`

=>
`I = 0.0007 41 \ A`