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A conducting rod of length 'l'=2sqrt(5) ...

A conducting rod of length `'l'=2sqrt(5)` meter and mass `'m'=4kg` lies on the horizontal table. Coefficient of friction between the rod and the table is `'mu'=(1/2)`. If the current in the conductor si 2 ampere, then find the minimum magnitude of magnetic field strength (in Tesla) such that conducting rod just starts to translate along `x` -axis (take `g=10m//sec^(2)`) [neglect the radius of rod]

Text Solution

Verified by Experts

The correct Answer is:
2
`F_("min")=(mumg)/(sqrt(1+mu^(2)))`
`il B_("min")=(mumg)/(sqrt(1+mu^(2)))`
`B_("min")=2` Tesla
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