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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.

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.

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Solution

The magnetic flux (Φ) through a surface is given by the dot product of the magnetic field (B) and the area vector (A). 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 (Tesla) A = length * width = 0.04 m * 0.05 m = 0.002 m² θ = 30° = 30 * π/180 rad = π/6 rad (conversion from degrees to radians)

Substituting these values into the formula:

Φ = 0.5 T * 0.002 m² * cos(π/6) Φ = 0.5 T * 0.002 m² * (√3/2) (since cos(π/6) = √3/2) Φ = 0.0005 Tm² * √3 Φ = 0.000866 T

So, the magnetic flux through the coil is approximately 0.000866 T*m² or 0.000866 Weber.

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