Question

A single-turn plane loop of wire with a cross-sectional area 200 cm2 is perpendicular to a magnetic field that increases uniformly from 0.200 T to 2.800 T in 2.20 seconds. What is the magnitude of the induced current if the resistance of the coil is 8.00 Ω?

Answer 1

Induced current,
`I=2.87* 10^(-3)\ A`

Step-by-step explanation:

Given that,

Area of cross section of the wire,
`A=200\ cm^2=0.02\ m^2`

Time, t = 2.2 s

Initial magnetic field,
`B_i=0.2\ T`

Final magnetic field,
`B_f=2.8\ T`

Resistance of the coil, R = 8 ohms

The expression for the induced emf is given by :

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

`\phi` = magnetic flux

`\epsilon=-(d(BA))/(dt)`

`\epsilon=A(d(B))/(dt)`

`\epsilon=A(B_f-B_i)/(t)`

`\epsilon=0.02* (2.8-0.2)/(2.2)`

`\epsilon=-0.023\ volts`

So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :

`\epsilon= IR`

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

`I=(0.023)/(8)`

`I=2.87* 10^(-3)\ A`

So, the magnitude of the induced current in the loop of wire is
`2.87* 10^(-3)\ A`. Hence, this is the required solution.