InterviewSolution
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InterviewSolution
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Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance `1Omega//m`. Position of the conducting rod at `t=0` is shown. A time dependent magnetic field `B=2t` tesla is switched on at `t=0` At `t=0`, when the magnetic field is switched on, the conducting rod is moved to the left at constant speed `5cm//s` by some external means. At `t=2s`, net induced emf has magnitudeA. (a) `0.12 V`B. (b) `0.08 V`C. ( c) `0.04 V`D. (d) `0.02 V` |
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Answer» 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. |
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