Question

a coil has 200 turns enclosing an area of 11.2 cm². in a physics laboratory experiment, the coil is rotated during the time interval 0.042 s from a position in which the plane of each turn is perpendicular to earths magnetic field to one in which the plane of each turn is parallel to the field. the magnitude of earths magnetic field at the lab location is 5.40×10⁻⁵ t. part a what is the magnitude |φ_initial| of the magnetic flux through one turn of the coil before it is rotated? express your answer in webers. view available hint(s) |φ_initial| = wb part b

Explanation:

Step1: Recall the formula for magnetic flux

The formula for magnetic flux is $\Phi = BA\cos\theta$, where $B$ is the magnetic field strength, $A$ is the area, and $\theta$ is the angle between the magnetic field and the normal to the area.

Step2: Determine the value of $\theta$ before rotation

Before rotation, the plane of the coil is perpendicular to the Earth's magnetic field. So, the angle $\theta$ between the magnetic field and the normal to the area is $0^{\circ}$, and $\cos\theta=\cos0^{\circ} = 1$.

Step3: Convert the area to SI units

The area $A = 11.2\ cm^{2}=11.2\times10^{- 4}\ m^{2}$ (since $1\ cm^{2}=10^{-4}\ m^{2}$).

Step4: Calculate the magnetic flux

Given $B = 5.40\times10^{-5}\ T$, using $\Phi = BA\cos\theta$ with $\cos\theta = 1$, we have $\Phi=(5.40\times 10^{-5}\ T)\times(11.2\times10^{-4}\ m^{2})\times1$.

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Answer:

$6.05\times 10^{-8}$ Wb