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

Answer:

The resistance of the wire is 5.15 ohm

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

$\Delta \phi _t=N\Delta \phi_1 =12\times 3.61\, Wb=43.32\, Wb$

Now induced voltage in the coil is given by

$\varepsilon _{ind}=\left |- \frac{\Delta \phi _t}{\Delta t} \right |={I_{ind}R$

where R = Resistance of the wire

$\therefore \varepsilon _{ind}=\left |- \frac{43.32}{0.0309} \right |\, volts=272\times R$

=>$R=5.15\, ohm$

Thus the resistance of the wire is 5.15 ohm