As shown below, a square loop of wire, 0.84 m on each side, has one edge along t
ID: 1744667 • Letter: A
Question
As shown below, a square loop of wire, 0.84 m on each side, has one edge along the positivez-axis and is tilted toward the y-zplane at an angle of 30.0^degree with respect to the horizontal(x-z plane). There is a uniform magnetic field of0.34 T pointing in the positivex-axis direction. (a) What is the flux through the loop? inWb (b) If the angle increases to 60^degree, what is the new flux throughthe loop? in Wb (c) While the angle is being increased, whichdirection will current flow through the top side of the loop?Explanation / Answer
Given that the side of the loop is L = 0.84m The magnetude of the magnetic field is B = 0.34T ------------------------------------------------------------------ The flux through the surface of the loop is =B.A = BA cos where is the angle between the area vector and the magneticfield ( In general area vectorperpendecular to the surface area that is normal to the surfacearea ) If the plane of the loop makesan angle = 300 then the angle betweenthe areal vector and the field is 600 when thefield makes 600 to the normal to theloop then the angle = 600 = BAcos = BA cos600 = B*4L *cos600 =---------- T where is the angle between the area vector and the magneticfield ( In general area vectorperpendecular to the surface area that is normal to the surfacearea ) If the plane of the loop makesan angle = 300 then the angle betweenthe areal vector and the field is 600 when thefield makes 600 to the normal to theloop then the angle = 600 = BAcos = BA cos600 = B*4L *cos600 =---------- T when thefield makes 600 to the normal to theloop then the angle = 600 = BAcos = BA cos600 = B*4L *cos600 =---------- T = BA cos600 = B*4L *cos600 =---------- T =---------- T If the plane of the loop makes an angle =600 then the angle between the areal vectorand the field is 300 when thefield makes 300 to the normal to theloop then the angle = 300 = BAcos = BA cos300 = B*4L *cos300 =---------- T From Righ hand rule the direction ofcurrent is counterclock wise . when thefield makes 300 to the normal to theloop then the angle = 300 = BAcos = BA cos300 = B*4L *cos300 =---------- T From Righ hand rule the direction ofcurrent is counterclock wise . when thefield makes 300 to the normal to theloop then the angle = 300 = BAcos = BA cos300 = B*4L *cos300 =---------- T =---------- T From Righ hand rule the direction ofcurrent is counterclock wise .Related Questions
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