1.

A square loop of length L is placed with its edges parallel to the XY-axies. The loop is carrying the current I. If the magnetic field in the region varies as B = B_(0) (1+(xy)/(L^(2)))hatk, then the magnitude of the force on the loop willl be

Answer»

`lB_(0)L`
`(lB_(0)L)/(2)`
` (lB_(0)L)/(sqrt2)`
`sqrt2 lB_(0)L`

Solution :`(**)`We take one vertex of square loop at origin,

for WIRE AB, magnetic field `AB=B_(0) (1+0) =B_(0),`
`therefore N =0`
For wire CD, magnetic fiels `= B_(0) (1+ (y)/(L)),`
`therefore x=L`
For wire BC, magnetic field `=B_(0) (1+ (x)/(L)),`
and for wire AD, magnetic field `=B_(0)`
`therefore y=0`
Now me calculate FORCES
`F_(1), F_(2), F_(3) and F_(4),`
`F_(1) = int _(0) ^(L) IB_(0) DY = IB_(0)L`
`F_(2)=int _(0) ^(I) IB _(0) (1+ (x)/(L)) dx=3/2 IB_(0)L`
`F_(3)- int _(0) ^(L) IB_(0) (1+ (y)/(L)) dx`
`= 3/2 IB_(0) L`
`F_(4) =int_(0) ^(L) IB_(0) dx = IB_(0)L`


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