Here, `r = 12 cm = 12 xx 10^(-2) m`,
`R = 8.5 ohm`
Induced current, `I = ?`
`A = pi r_(2) = (22)/(7) (12 xx 10^(-2))^(2) = 4.52 xx 10^(-2) m^(2)`
From the given graph
`(dB)/(dt) = (1)/(2) T//s` for `t = 0` to `t = 2 s`
`(dB)/(dt) = 0` for `t = 2` to `t = 4 s`
`(dB)/(dt) = - (1)/(2) T//s` for `t = 4 s` to `t = 6 s`
`i = (e)/(R) = (d phi // dt)/(R) = (A(dB // dt))/(R)`
For `t = 0` to `t = 2 s`
`i = (4.52 xx 10^(-2))/(8.5) xx (1)/(2) = - 0.266 xx 10^(-2) A`
` = - 2.66 mA`
For `t = 2s` to `t = 4s`
`i = 0`
For `t = 2s` to `t = 4s`
For `t = 4s` to `t = 6s`
`i = - (4.52 xx 10^(-2) (-(1)/(2)))/(8.5) = + 2.66 mA`
The graphical variation of induced current with time is shown in fig. From `t = 0` to `t = 2s`, magnetic field is increasing. Therefore, induced current oppose the increase. From `t = 2s` to `t = 4s` induced current is zero. From `t = 4s` to `t = 6s` magnetic field is decreasing. Therefore, induced current opposes the decrease and flow n the same direction.
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/16109388592399210391610938859.png)