Question

A 19.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.8-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.27 s .

What is the average induced emf in the loop?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer 2

To find the average induced emf in the loop, we'll use Faraday's Law of Induction:

$$\text{emf}=-\frac{\mathrm{\Delta }{\mathrm{\Phi }}_B}{\mathrm{\Delta }t}$$

Step 1: Calculate the initial and final magnetic flux.

The magnetic flux (${\mathrm{\Phi }}_B$) through the loop is given by:

$${\mathrm{\Phi }}_B=B\cdot A\cdot \cos \theta$$

where:

• $B=1.8\text{T}$ (magnetic field),

• $A$ is the area of the loop,

• $\theta$ is the angle between the magnetic field and the normal to the loop.

Initial flux (${\mathrm{\Phi }}_{\text{initial}}$):
The loop is initially perpendicular to the field, so $\theta =0^{\circ }$ and $\cos 0^{\circ }=1$.

$${\mathrm{\Phi }}_{\text{initial}}=B\cdot A\cdot 1=1.8\cdot \pi {\left(\frac{0.195}{2}\right)}^2$$$${\mathrm{\Phi }}_{\text{initial}}=1.8\cdot \pi \cdot (0.0975{)}^2\approx 1.8\cdot \pi \cdot 0.009506\approx 0.0537\text{Wb}$$

Final flux (${\mathrm{\Phi }}_{\text{final}}$):
The loop is rotated to be parallel to the field, so $\theta =90^{\circ }$ and $\cos 90^{\circ }=0$.

$${\mathrm{\Phi }}_{\text{final}}=B\cdot A\cdot 0=0\text{Wb}$$

Step 2: Calculate the change in flux ($\mathrm{\Delta }{\mathrm{\Phi }}_B$).

$$\mathrm{\Delta }{\mathrm{\Phi }}_B={\mathrm{\Phi }}_{\text{final}}-{\mathrm{\Phi }}_{\text{initial}}=0-0.0537=-0.0537\text{Wb}$$

Step 3: Calculate the average induced emf.
The time taken ($\mathrm{\Delta }t$) is $0.27\text{s}$.

$$\text{emf}=-\frac{\mathrm{\Delta }{\mathrm{\Phi }}_B}{\mathrm{\Delta }t}=-\left(\frac{-0.0537}{0.27}\right)\approx 0.20\text{V}$$