Answer:
The induced emf in the coil at the t = 5s is 6.363 mV
Explanation:
Given;
number of turns = 75
diameter of the coil = 6 cm
magnetic field strength = 1 T
new magnetic field strength = 1.30 T at t = 10.0 s
[tex]Area \ of \ coil = \frac{\pi d^2}{4} = \frac{\pi *0.06^2}{4} = 0.002828 \ m^2[/tex]
[tex]E.M.F = \frac{NA* \delta B}{\delta t}[/tex]
Between 0 to 5 s, Induced emf is given as;
[tex]E.M.F = \frac{75*0.002828*(B_5-1)}{5}[/tex]
Between 5 to 10 s, Induced emf is given as;
[tex]E.M.F = \frac{75*0.002828*(1.3-B_5)}{5}[/tex]
Since the field increased at a uniform rate until it reaches 1.30 T at t = 10.0 s, the induced emf will also increase in uniform rate. And equal time interval will generate same increase in field strength.
B₅ -1 = 1.3 - B₅
2B₅ = 2.3
B₅ = 1.15 T
Thus, magnetic field at t = 5 is 1.15 T
[tex]E.M.F = \frac{75*0.002828*(1.3-1.15)}{5} = 6.363 \ mV[/tex]
Therefore, the induced emf in the coil at the t = 5s is 6.363 mV
