Answer:
Angular speed of the coil 5.08 rad/s
Solution:
As per the solution:
No. of turns, N = 150 turns
Length of the coil, l = 2.00 cm
Magnetic field, B = [tex]8.20\times 10^{- 2}\ T[/tex]
Maximum emf, [tex]e_{max} = 2.50\times 10^{- 2}[/tex]
Now,
[tex]\phi = BAcos\omega t[/tex]
Instantaneous emf, [tex]e = -N\frac{d\phi}{dt}[/tex]
where
[tex]\phi [/tex] = flux
[tex]\omega [/tex] = angular speed
[tex]e = -N\frac{d(BAcos\omega t)}{dt}[/tex]
[tex]e = N\omega (BAsin\omega t)[/tex]
Maximum emf is obtained at [tex]sin90^{\circ}[/tex]
[tex]e_{max} = BAN\omega [/tex]
[tex]\omega = \frac{e_{max}}{BAN}[/tex]
[tex]\omega = \frac{2.50\times 10^{- 2}}{8.20\times 10^{- 2}\times 0.02^{2}\times 150} = 5.08\ rad/s[/tex]
