A uniform magnetic field, B of strength 0.5T is directed at an angle of 30° to the plane of a rectangular copper coil of length 0.04m and width 0.05m. Determine the flux through the coil.*1 point0.00050 Wb0.00200 Wb0.00400 Wb0.00866 Wb
Question
A uniform magnetic field, B of strength 0.5T is directed at an angle of 30° to the plane of a rectangular copper coil of length 0.04m and width 0.05m. Determine the flux through the coil.*1 point0.00050 Wb0.00200 Wb0.00400 Wb0.00866 Wb
Solution
The magnetic flux (Φ) through a surface is given by the dot product of the magnetic field (B) and the area vector (A) of the surface. The dot product takes into account the angle (θ) between the two vectors. The formula is:
Φ = B * A * cos(θ)
Given: B = 0.5 T, A = length * width = 0.04 m * 0.05 m = 0.002 m², θ = 30° = π/6 rad (converted from degrees to radians because the trigonometric functions in the formula use radians).
Substituting these values into the formula:
Φ = 0.5 T * 0.002 m² * cos(π/6 rad) = 0.5 T * 0.002 m² * (√3/2) = 0.0005 T*m² = 0.0005 Wb.
So, the magnetic flux through the coil is 0.0005 Wb.
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