00:01
Here first we calculate b vector, means the magnetic field density.
00:05
And we know b vector is equal to curl of a vector.
00:12
Here we are solving part a.
00:15
Now we know curl of a vector is equal to i, z, k, del over del x, del over del y, del over del z, coefficient of i is equal to 2 x square y plus y z coefficient of z coefficient of z x y square minus x z cube coefficient of k 6 x y z minus 2 x square y square now this becomes equals to i, del over del y, 2x square, y square minus 6x y z minus del over day z x y square minus x y squared minus j, del over del x squared minus j, del over del x squared, minus j, del over del x of two ox square, y square minus 6 x y z minus del over del j z two x square y plus y plus k del over del del x of x y square minus x x cube minus del over del y of two x square y plus y plus y z now by simplifying this we obtain b vector is equal to i the partial derivative of 2 x square y square minus 6 y z with respect to y comes equal to 4 x square y minus 6 x z and the del over del z of this term comes equal to negative 3 x z square minus z of this term with respect to x comes equal to 4x y square minus 6 y z minus the partial derivative of this term with respect to z comes to equal to y plus k the partial derivative of this term with respect to x comes equal to y square minus z cube minus the partial derivative of this term with respect to y comes equal to 2 x square plus z.
03:22
Now by simplifying this we obtain b vector is equal to 4 x square y minus 6 x x plus x plus 3 x x squared i plus 6 y z plus y minus 4 x x x square z plus y minus 4 x y square z plus z plus y cube minus z cube minus two x square minus z k now next we find the magnetic flux passing through a loop described by x equals to 1 and y z from 0 to 2 and we know magnetic flux 5 is given by double integration b vector dot d s vector here x equals to 1 that means d x x comes equal to 0.
04:18
And we know ds vector is equal to dy, dz, i, plus dx, dz, j, plus dx, dz, j, plus dx, dx, dy, k.
04:33
And here because dx is equal to zero, so from this we obtain d .s vector is equal to dy, dz, i.
04:43
Now next we find the dot product of b vector and ds vector.
04:46
And here this is b vector so b vector dot ds ds vector comes equal to 4 x square y minus 6 x z plus 3 x z square b y d z and other terms becomes equal to 0 so from this we obtained phi is equal to double integration 4 x square y minus 6 x z z plus 3 x z square dy d z here the value of x is equal to 1 so this becomes equal to 4 y minus 6 z plus 3 z square dy d z here z varies from 0 to 2 and y also varies from 0 to 2 because here y and z are lies between 0 to 2 now after integrating this expression with respect to z, this becomes equal to 0 to 2, 4, y, z minus 3, z squared, 0 to 2, b ,y.
06:04
Now after putting upper limits and low limits, this becomes equal to 8y minus 12 plus 8 by...