Question

The magnetic flux that passes through one turn of a 12-turn coil of wire changes to 5.50 from 9.11 Wb in a time of 0.0309 s. The average induced current in the coil is 272 A. What is the resistance of the wire?

Answer 1

The resistance of the wire is 5.15 ohm

Step-by-step explanation:

Given Magnet flux change through one turn is

`\Delta \phi_1 =\left | \phi _f-\phi _i \right |=\left | 9.11-5.50 \right |Wb=3.61\, Wb`

Time take
`\Delta t=(0.0309-0)\, s=0.0309\, s`

Induced current
`I_(ind)=272\, A`

Number of turns ,N= 12 turns

Therefore total change in flux through 12 turns is

`\Delta \phi _t=N\Delta \phi_1 =12* 3.61\, Wb=43.32\, Wb`

Now induced voltage in the coil is given by

`\varepsilon _(ind)=\left |- (\Delta \phi _t)/(\Delta t) \right |={I_(ind)R`

where R = Resistance of the wire

`\therefore \varepsilon _(ind)=\left |- (43.32)/(0.0309) \right |\, volts=272* R`

=>
`R=5.15\, ohm`

Thus the resistance of the wire is 5.15 ohm