Correct Answer - B
(b) `(dB)/(dt) = 2 T//s`
`E = -(AdB)/(dt) = -800 xx 10^(-4)m^(2) = 0.16 V`
`I = (0.16)/(1 Omega) = 0.16 A, clockwise`
At `t = 2 s, B = 4 T, (dB)/(dt) = 2 T//s `
`a = 20 xx 30 cm^(2)`
`= 600 xx 10^(-4)m^(2), (dA)/(dt) = -(5 xx 20)cm^(2)//s`
`= -100 xx 10^(-4)m^(2)//s`
`E = -(dphi)/(dt) = -[(d(BA))/(dt)] = -[(BdA)/(dt) + (AdB)/(dt)]`
`= -[4 xx(-100 xx 10^(-4)) + 600 xx 10^(-4) xx 2]`
`= -[-0.04 + 0.120] = -0.08 V`
Alternative: `phi = BA = 2t xx 0.2(0.4 - vt)`
`E = -(dphi)/(dt) = 0.8 vt - 0.16`
At `t = 2s`
`E = 0.08 V`
At `t =2s`, "length of the wire" `= (2 xx 30 cm) + 20cm = 0.8`
Resistance of wire `= 0,8 Omega`
Current through the rod `= (0.08)/(0.8) = (1)/(10)A`
Force on the wire is `= ilB`
`= (1)/(10) xx (0.2) xx 4 = 0.08 N`
Same force is placed on the rod in opposite direction to make net force zero.