Acircular coil 17.1 cm in diameter and containing 11 loops lies flat on the gound. the earth's magnetic field at this location has magnitude 5.5e-5 t and points into the earth at an angle of 66.0 degrees below the horizontal line and aimed due north. determine the torque if a 7.48 a clockwise current passes through the coil.

Answer

given that

magnitude of magnetic field B = 5.5 x 10⁻⁵T

diameter of coil = d = 17.1cm

                                =0.171 m

radius of the coil r = d/2

                               = 0.0855 m

area of the loop A = πr²

                              = ( 3.14 ) ( 0.0855 m)²

                              =0.0229 m²

(a) The angle between the normal to the coil and the field is 34.0°, so the torque is

                   q = 90 - 66 = 24°

                   t = N I A B sin q  

                    = (11) (7.48 A) (0.0229 m²) (5.50 x 10⁻⁵T) sin 24.0°  

                   =   4.215 x 10⁻⁵ N. m

change tension

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid