A straight conductor of weight 1 N and length 0.5 m, is located in a plane making an angle `30^@` with the horizontal so that it is perpendicular to a uniform mangetic field of induction `B= 0.109 T.` Given that the conductor carries a current of 10 A away from the reader, and coefficient of static friction is 0.1, find the force needed to be applied parallel to the plane to sustain the conductor at rest.

Answer

` 0. 62 N < F < 0.88 N`

Explanation

`F_(r) =muN =mu(mg +ilB) cos30^(@)`
`F_(min) = (mg + IlB) sin 30^(@) - mu (mg + IlB) cos 30^(@)`
`F_(min) = (1 -10 xx 0.5 xx 0.1) xx (1)/(2) - 0.1 (1 + 10 xx 0.5 xx 0.1) xx (sqrt3)/(2)`
`F_(min) = 0.75 - 0.1 xx 0.75 xx 0.866 = 0.62 N`
`F_(max) = (mg + IlB) sin 30^(@) + mu (mg + IlB) cos 30^(@)`
` =0.88 N`
`0.62 N le 0.88 N`
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