Correct Answer - `-(mu_(00I))/(4R) q v_(0) hatk` (ii) `F_(1) = 2I R B F_(2) = 2I R`
`B` Net Force ` =F_(1) + F_(2) = 4 I R B hati` .
`B_(1)` = field due to `KNM`
`B_(2) =` field due to to `KLM`
field at origin `oversetrarr(B_(0)) = oversetrarr(B_(1))oversetrarr(B_(2))`
` = (mu_(0))/(4pi) (piI)/(R ) hatj + (mu_(0))/(4pi) (-hati) = (mu_(0))/(4) (I)/R (-hati +hatj)`
Force on particle
`oversetrarrF = q (oversetrarrV xx oversetrarrB) = q V_(0) B [-hati xx (-hati +hatj)]`
`gV_(0) B_(0) (-hatk) = (mu_(0)I)/(4R) qV_(0) (-hatk)`
(ii) In a uniiform field the force on the curved loop is `oversetrarrF = i (oversetrarrl xx oversetB)` where `vecl =rarr ` vector from one end of loop to the other end
`:. oversetrarrF_(KLM) = i (KoversetrarrM xx oversetrarrB) i [2R (-hatk)xx hatB hatj] =2BiR hati`
`:.oversetrarrF = oversetrarrF_(KLM) + oversetrarrF_(KLM) =4BiRhati` .
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/16110713252148794631611071325.png)
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