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A metal rod of mass 10 gm and length 25 cm is suspended on two springs as shown in Fig. 1.131. The springs are extended by 4 cm. When a 20 A current passes through the rod, it rises by 1 cm. Determine the magnetic field assuming acceleration due to gravity to be `10 ms^-1.` |
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Answer» Correct Answer - `1.5xx10^-3T` Initially, the rod will be in equilibrium if `2T_0=mg with T_0=kx_0....(i)` When current is passed through the rod, it experiences a force `F=Bil` vertically upward. Now for equilibrium, in this situation `2T+Bil=mg with T=kx......(ii)` From equations (i) and (ii), we get `T/(T_0)=(mg-Bil)/(mg) or x/(x_0)=1-(Bil)/(mg)` Solve to get `B=(mg(x_0-x))/(ilx_0)=((1xx10^-3)(10)(3xx10^-2))/((20)(25xx10^-2)(4xx10^-2))` `=1.5xx10^-3T` |
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