Confused with which value to use in my equation?

Question

One of my assignment question states that a ring lies on the horizontal, the earths magnetic field is acting 66 degrees below the horizontal. Find the value of the magnetic flux. I know that the equation is ABcos(angle), I have the values for A and B it is the cos(angle) that is causing me problems, is the value for angle 66 degrees or 24 degrees

Answer 1

You seem to know the formula for finding magnetic flux as A.B i.e. it is the dot product of the Area vector and the Magnetic field vector. Remember that the area vector is ALWAYS normal to the plane of the surface you are dealing with. So in this case we have to simply get the component of the field PERPENDICULAR to the ring. Since the field is directed at an angle of 66 deg. BELOW THE HORIZONTAL and the ring is horizontal so we have the field making an angle of 66deg. with the ring's plane surface and angle of 24deg. with the normal (line perpendicular to the ring) to the ring. So you have to use the value of cos 24 which is 0.913 and multiply it with the magnitudes of the ring's area and the field's strength. Contact me if you have any doubts or queries regarding this topic or dot product.

Answer 2

You want the part of the B field that "goes through" the area.  If the ring is "flat", i.e., open area UP toward the sky, then you want cos(24)...the amount of B that will "cut through" the ring's area.

If the ring sits on the horizontal with its area open in the "East-West" direction, then you want cos(66).

-Fred

Answer 3

cos(66) is not equal to cos(24).For odd multiples of 90

              sine gets transformed to cosine

              secant gets transformed into cosecant

               tangent gets transformed into cotangent

                

Here cos is getting transformed into cos itself.so the value of

  

                 cos(66) is not equal to cos (24)

                   value of cos(66)=value of sin(24)

Answer 4

It is EXACTLY how Getfred said.  Just do as he explained.  The guy under him agrees with his answer but not as clearly, in my opin.

Answer 5

The flux of a vector field B through a surface of area A is Φ = B.A.cosα, where α is the angle made by vector B and the normal to the surface. If vector B is 66° below the horizontal, it will make with the normal an angle of 24° (the ring is horizontal). 24°  is the value you should use.