Question

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.`

Answer 1

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`