00:01
Here we have a diagram of our rod which makes an angle of 36 .9 degrees to the positive x -axis and 53 .1 degrees to the positive y -axis.
00:12
Since the bar is straight and the mingering field is uniform, if we integrate the epsilon is equal to v cross v .dl along the length of the bar, we get that epsilon is equal to v cross b dotted with the length vector l.
00:27
So what do we know? firstly for the bar we know that the velocity vector v is equal to 4 .2 meters per second and that's in the eye direction.
00:45
L the length vector of the rod will break up into its x and y components since the rod lies diagonally with respect to our axes so the length of the rod is 0 .25 meters multiplied by costs of 36 .9 degrees in the i direction plus sign of 36 .9 degrees in the j direction so that's the y component of the length of the rod so now we have v and l we are also given b so we can calculate the emf induced in the rod so the emph epsilon is equal to v cross b and then the result in vector dotted with the limb vector l.
01:56
And so performing this, putting in our values, we get at 4 .2 meters the second high crossword, magnetic field, which is 0 .1, 2 teslas in the direction minus 0 .22 teslas in the j direction in the j direction, in the j direction, and direction minus 0 .09, 0 .09, 0 .0 .0 .0 .0 .0 .0 .0 .0 .8 0 .0 .0 .8 .8.
03:07
Vols per meter, jade direction.
03:45
Minus 0 .924 volts per meter multiplied by the unit vector k and this is dotted with 0 .25 meters...