Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
![\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdelta%20B%7D%7B%5Cdelta%20t%7D%20%3D%20%5Cfrac%7B%5Cmu%20N%20%5Cfrac%7B%5Cdelta%20i%7D%7B%5Cdelta%20t%7D%20%7D%7BL%7D%20%5C%5C%5C%5C%5Cfrac%7B%5Cdelta%20B%7D%7B%5Cdelta%20t%7D%20%3D%20%5Cfrac%7B4%5Cpi%20%2A10%5E%7B-7%7D%20%2A600%2A%20%5Cfrac%7B5%7D%7B0.6%7D%20%7D%7B0.25%7D%5C%5C%5C%5C%5Cfrac%7B%5Cdelta%20B%7D%7B%5Cdelta%20t%7D%20%3D0.02514%20%5C%20T%2Fs)
The induced emf in the shorter coil is calculated as;
![E = NA\frac{\delta B}{\delta t}](https://tex.z-dn.net/?f=E%20%3D%20NA%5Cfrac%7B%5Cdelta%20B%7D%7B%5Cdelta%20t%7D)
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
![E = NA\frac{\delta B}{\delta t}](https://tex.z-dn.net/?f=E%20%3D%20NA%5Cfrac%7B%5Cdelta%20B%7D%7B%5Cdelta%20t%7D)
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V