A 200-loop coil of cross sectional area 8.5 lies in the plane of the paper. Directed out of the plane of the paper is a magnetic field of 0.06 T. The field out of the paper decreases to 0.02 T in 12 milliseconds.
(a) What is the magnitude of the change in magnetic flux enclosed by the coil?
To find the magnitude of the change in magnetic flux enclosed by the coil, we need to calculate the initial and final magnetic flux and then subtract them.
The magnetic flux (Φ) is the product of the magnetic field (B) and the area (A) of the coil:
Φ = B * A
Given:
Number of loops (N) = 200
Cross-sectional area (A) = 8.5 m²
Initial magnetic field (B₁) = 0.06 T
Final magnetic field (B₂) = 0.02 T
Time taken (Δt) = 12 milliseconds = 12 * 10⁻³ s
To calculate the initial magnetic flux (Φ₁), substitute the given values into the formula:
Φ₁ = B₁ * A
Φ₁ = 0.06 T * 8.5 m²
To calculate the final magnetic flux (Φ₂), substitute the given values into the formula:
Φ₂ = B₂ * A
Φ₂ = 0.02 T * 8.5 m²
Now, to calculate the magnitude of the change in magnetic flux (ΔΦ), we subtract the initial flux from the final flux:
ΔΦ = Φ₂ - Φ₁
Finally, substitute the values:
ΔΦ = (0.02 T * 8.5 m²) - (0.06 T * 8.5 m²)
Simplify the expression:
ΔΦ = (0.17 - 0.51) T * m²
ΔΦ = -0.34 T * m²
Therefore, the magnitude of the change in magnetic flux enclosed by the coil is 0.34 T * m².
change in flux= .04T
dflux/dt=.02/.012