AP Physics 2 Exam Question:

A rectangular loop of wire with dimensions 0.1 m by 0.2 m is placed in a region where a uniform magnetic field of 0.5 T is directed into the page. The loop is initially oriented with its longer side parallel to the direction of the magnetic field.

a) Calculate the magnetic flux through the loop. b) If the loop is rotated by 90 degrees such that its longer side is now perpendicular to the magnetic field, calculate the change in magnetic flux through the loop. c) Determine the magnitude and direction of the induced emf in the loop when it is rotated by 90 degrees in the given magnetic field.

Answer:

a) Magnetic flux ($\Phi$) through a loop is given by the equation:

$\Phi = B \cdot A \cdot \cos(\theta)$

Where:

• B is the magnetic field strength,
• A is the area of the loop,
• $\theta$ is the angle between the magnetic field and the normal to the loop.

In this case, the magnetic field is directed into the page, and the loop is initially oriented with its longer side parallel to the field. Thus, $\theta = 0$, and $\cos(\theta) = \cos(0) = 1$.

The area of the loop can be calculated as:

$A = \text{length} \cdot \text{width} = 0.1 \, \text{m} \cdot 0.2 \, \text{m} = 0.02 \, \text{m}^2$

Substituting the given values into the formula for magnetic flux:

$\Phi = (0.5 \, \text{T}) \cdot (0.02 \, \text{m}^2) \cdot 1 = 0.01 \, \text{T} \cdot \text{m}^2$

Thus, the magnetic flux through the loop is 0.01 T·m².

b) When the loop is rotated by 90 degrees, the angle $\theta$ becomes 90 degrees, and $\cos(\theta) = \cos(90) = 0$.

The change in magnetic flux ($\Delta \Phi$) through the loop can be calculated as:

$\Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}}$

Since $\cos(\theta)$ changes from 1 to 0, the initial magnetic flux through the loop is:

$\Phi_{\text{initial}} = (0.5 \, \text{T}) \cdot (0.02 \, \text{m}^2) \cdot 1 = 0.01 \, \text{T} \cdot \text{m}^2$

And the final magnetic flux through the loop is:

$\Phi_{\text{final}} = (0.5 \, \text{T}) \cdot (0.02 \, \text{m}^2) \cdot 0 = 0 \, \text{T} \cdot \text{m}^2$

Thus, the change in magnetic flux through the loop is:

$\Delta \Phi = 0 \, \text{T} \cdot \text{m}^2 - 0.01 \, \text{T} \cdot \text{m}^2 = -0.01 \, \text{T} \cdot \text{m}^2$

c) According to Faraday's Law of electromagnetic induction, the magnitude of the induced electromotive force (emf) in the loop is equal to the rate of change of magnetic flux through the loop. Mathematically, it is given by:

$\epsilon = -\dfrac{d\Phi}{dt}$

In this case, the change in magnetic flux through the loop when it is rotated by 90 degrees is $\Delta \Phi = -0.01 \, \text{T} \cdot \text{m}^2$. The time taken for the rotation is not given, but let's assume it takes 1 second for simplicity.

Thus, the magnitude of the induced emf in the loop is:

$\epsilon = -\dfrac{\Delta \Phi}{dt} = -\dfrac{-0.01 \, \text{T} \cdot \text{m}^2}{1 \, \text{s}} = 0.01 \, \text{V}$

The negative sign indicates that the direction of the induced emf opposes the change in magnetic flux. Since the flux is decreasing (from 0.01 T·m² to 0 T·m²), the induced emf is in the clockwise direction within the loop.