Electromagnetic FieldsThis revised edition provides patient guidance in its clear and organized presentation of problems. It is rich in variety, large in number and provides very careful treatment of relativity. One outstanding feature is the inclusion of simple, standard examples demonstrated in different methods that will allow students to enhance and understand their calculating abilities. There are over 145 worked examples; virtually all of the standard problems are included. |
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Page 38
... Given the two vectors A - = 1-4 ✰ + 2ŷ + 32 and B = 4 – 5ŷ + 62 , find the angle between them . Find the component of A in the direction of B. = - 6x ---- = Given the vectors A 2x + 3ŷ - 42 and 4ŷ + 2 . Find the component of A X B ...
... Given the two vectors A - = 1-4 ✰ + 2ŷ + 32 and B = 4 – 5ŷ + 62 , find the angle between them . Find the component of A in the direction of B. = - 6x ---- = Given the vectors A 2x + 3ŷ - 42 and 4ŷ + 2 . Find the component of A X B ...
Page 39
... given by : ( 0 , 0 ) → ( 3,0 ) - > ( 3 , 4 ) → ( 0 , 4 ) ( 0 , 0 ) . Also calculate the surface integral of ▾ A over the enclosed area and show that ( 1-67 ) is satisfied . 1-15 Given the vector field A = x2y + xy2ŷ + a3e By cos ax2 ...
... given by : ( 0 , 0 ) → ( 3,0 ) - > ( 3 , 4 ) → ( 0 , 4 ) ( 0 , 0 ) . Also calculate the surface integral of ▾ A over the enclosed area and show that ( 1-67 ) is satisfied . 1-15 Given the vector field A = x2y + xy2ŷ + a3e By cos ax2 ...
Page 275
... given by ( 17-3 ) . We find the flux in the same manner as we used to get ( 17-14 ) : = В S B. în da = Boab cos = Boab cos ( wt + 0 ) ( 17-36 ) & ind = wBoab sin ( wt + Po ) ( 17-37 ) Substituting this into ( 17-3 ) , we find the ...
... given by ( 17-3 ) . We find the flux in the same manner as we used to get ( 17-14 ) : = В S B. în da = Boab cos = Boab cos ( wt + 0 ) ( 17-36 ) & ind = wBoab sin ( wt + Po ) ( 17-37 ) Substituting this into ( 17-3 ) , we find the ...
Contents
INTRODUCTION | 1 |
ELECTRIC MULTIPOLES | 8 |
THE VECTOR POTENTIAL | 16 |
Copyright | |
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Ampère's law angle assume axes axis bound charge boundary conditions bounding surface calculate capacitance charge density charge distribution charge q circuit conductor consider const constant corresponding Coulomb's law curve cylinder dielectric dipole direction distance divergence theorem E₁ electric field electromagnetic electrostatic energy equation evaluate example expression field point free charge function given induction infinitely long integral integrand Laplace's equation line charge line integral located magnetic magnitude Maxwell's equations obtained origin P₁ perpendicular point charge polarized position vector potential difference quadrupole R₁ region result scalar potential Section shown in Figure sphere of radius spherical surface charge surface charge density surface integral tangential components theorem total charge vacuum vector potential velocity volume wave write written xy plane zero Απερ дх