an 8.8-m length of conducting wire is formed into a square and placed in a unifo

Question

an 8.8-m length of conducting wire is formed into a square and placed in a uniform magnetic field strength of 1.2 T. the square is then rotated continuously in three dimensions, so that the angle ap between the normal line to the square and the magnetic field varies between 0 degrees and 90 degrees. at what angle ap does the magnetic flux through the conductor equal 1.4 T*m^2? an 8.8-m length of conducting wire is formed into a square and placed in a uniform magnetic field strength of 1.2 T. the square is then rotated continuously in three dimensions, so that the angle ap between the normal line to the square and the magnetic field varies between 0 degrees and 90 degrees. at what angle ap does the magnetic flux through the conductor equal 1.4 T*m^2? an 8.8-m length of conducting wire is formed into a square and placed in a uniform magnetic field strength of 1.2 T. the square is then rotated continuously in three dimensions, so that the angle ap between the normal line to the square and the magnetic field varies between 0 degrees and 90 degrees. at what angle ap does the magnetic flux through the conductor equal 1.4 T*m^2?

Explanation / Answer

The magnetic flux is defined as:

F = BA cos(q) where q is the angle between the normal to the area A and the magnetic field direction

Youe are given F = 1.4 T-m^2. YOu are told that a wire of length L = 8.8 m is formed into a square. Since all four sides of a square are of the same length, one side is just:

s = L/4 = 2.2 m

in length. The area A is then:

A = s^2 = (L/4)^2 = (2.2 m)^2 = 4.84 m^2

Now you can solve the flux equation for q:

cos(q) = F/(BA) ---> q = arccos(F/(BA)) = arccos(1.4 T-m^2/(1.2 T* 4.84 m^2)) = arccos(0.24105)

q = 76.05 deg

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