Question

The magnetic field through a circular loop of wire 12 cm in radius and 8.5 ohm resistance, changes with time as shown in Fig. The magnetic field is perpendicular to the plane of the loop. Calculate the induced current in the loop and plot is as a function of time.

Answer 1

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.