A) 0.25
B) 0.31
C) 0.43
D) 0.12
Correct Answer: D
Solution :
The current through the wire PQ is \[I=\varepsilon /R\] When the wire just starts sliding, the force \[I\ell B\] toward right is equal to the force \[\mu \] mg towards left. \[\Rightarrow \,\,\mu mg\,=I\ell B\,=(\varepsilon R)\ell B\] \[\therefore \,\,\mu =\,\frac{\varepsilon \ell B}{mgR}\,=\frac{6\times 4.9\times {{10}^{-2}}\,\times 0.8}{{{10}^{-2}}\,\times 9.8\,\times 20}\,=0.12\]You need to login to perform this action.
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