Question

You are looking down on a single coil in a constant magnetic field B = 1.2 T which points directly into of the screen. The dimensions of the coil go from a = 8 cm and b = 17 cm, to a* = 16 cm and b* = 22 cm in t = 0.04 seconds. If the coil has resistance that remains constant at 1.2 ohms. What would be the magnitude of the induced current in amperes?

Answer 1

The current is
`I = 0.5425 \ A`

Step-by-step explanation:

From the question we are told that

The magnetic field is
`B = 1.2 \ T`

The first length is
`a = 8 \ cm = 0.08 \ m`

The second length is
`a^* = 16 \ cm = 0.16 \ m`

The first width is
`b = 17 \ cm = 0.17 \ m`

The second width is
`b^* = 22 \ cm = 0.22 \ m`

The time interval is
`dt = 0.04 \ s`

The resistance is
`R = 1.2 \ \Omega`

Generally the first area is

`A = a * b`

=>
`A = 0.08 * 0.17`

=>
`A = 0.0136 \ m^2`

The second area is

`A^* = a^* * b^*`

=>
`A^* = 0.16 * 0.22`

=>
`A^* = 0.0352 \ m^2`

Generally the induced emf is mathematically represented as

`\epsilon = - ( B * [A^* - A])/(dt)`

This negative show that it is moving in the opposite direction of the motion producing it

=>
`|\epsilon | = ( 1.2 * [ 0.0352-0.0135])/(0.04)`

=>
`|\epsilon | = 0.651 \ V`

The induced current is

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

=>
`I = ( 0.651)/(1.2)`

=>
`I = 0.5425 \ A`