A student takes a strand of copper wire and forms it into a circular loop with a circumference of 0.335 m. The student then places the loop in a uniform, constant magnetic field of magnitude 0.00313 T that is oriented perpendicular to the face of the loop. Pulling on the ends of the wire, they reduce the circumference of the loop to 0.167 m in a time interval of 0.682 s. Assuming that the loop remains circular as it shrinks, what is the magnitude of the average emf induced around the loop during this time interval?
The Correct Answer and Explanation is:
To calculate the magnitude of the average induced electromotive force (emf) around the loop, we can apply Faraday’s Law of Induction. This law states that the induced emf is equal to the rate of change of the magnetic flux through the loop:E=−ΔΦΔt\mathcal{E} = – \frac{\Delta \Phi}{\Delta t}E=−ΔtΔΦ
where:
- E\mathcal{E}E is the induced emf,
- ΔΦ\Delta \PhiΔΦ is the change in magnetic flux,
- Δt\Delta tΔt is the time interval over which the change occurs.
Step 1: Magnetic Flux
The magnetic flux (Φ\PhiΦ) through the loop is given by the formula:Φ=B⋅A\Phi = B \cdot AΦ=B⋅A
where:
- BBB is the magnetic field strength (0.00313 T),
- AAA is the area of the loop.
Since the loop is circular, the area AAA can be calculated using the formula:A=πr2A = \pi r^2A=πr2
where rrr is the radius of the loop. The radius can be derived from the circumference CCC, using the formula C=2πrC = 2\pi rC=2πr. Therefore, the radius is:r=C2πr = \frac{C}{2\pi}r=2πC
For the initial circumference C1=0.335 mC_1 = 0.335 \, \text{m}C1=0.335m, the initial radius is:r1=0.3352π≈0.0533 mr_1 = \frac{0.335}{2\pi} \approx 0.0533 \, \text{m}r1=2π0.335≈0.0533m
Now, the initial area A1A_1A1 is:A1=π(0.0533)2≈0.00896 m2A_1 = \pi (0.0533)^2 \approx 0.00896 \, \text{m}^2A1=π(0.0533)2≈0.00896m2
For the final circumference C2=0.167 mC_2 = 0.167 \, \text{m}C2=0.167m, the final radius is:r2=0.1672π≈0.0266 mr_2 = \frac{0.167}{2\pi} \approx 0.0266 \, \text{m}r2=2π0.167≈0.0266m
Now, the final area A2A_2A2 is:A2=π(0.0266)2≈0.00222 m2A_2 = \pi (0.0266)^2 \approx 0.00222 \, \text{m}^2A2=π(0.0266)2≈0.00222m2
Step 2: Change in Magnetic Flux
The change in magnetic flux ΔΦ\Delta \PhiΔΦ is:ΔΦ=B⋅(A2−A1)\Delta \Phi = B \cdot (A_2 – A_1)ΔΦ=B⋅(A2−A1)
Substituting the values:ΔΦ=0.00313⋅(0.00222−0.00896)≈0.00313⋅(−0.00674)≈−2.11×10−5 Wb\Delta \Phi = 0.00313 \cdot (0.00222 – 0.00896) \approx 0.00313 \cdot (-0.00674) \approx -2.11 \times 10^{-5} \, \text{Wb}ΔΦ=0.00313⋅(0.00222−0.00896)≈0.00313⋅(−0.00674)≈−2.11×10−5Wb
Step 3: Calculate the Induced emf
The time interval over which the change occurs is Δt=0.682 s\Delta t = 0.682 \, \text{s}Δt=0.682s. Now we can calculate the induced emf:E=−ΔΦΔt=2.11×10−50.682≈3.09×10−5 V\mathcal{E} = – \frac{\Delta \Phi}{\Delta t} = \frac{2.11 \times 10^{-5}}{0.682} \approx 3.09 \times 10^{-5} \, \text{V}E=−ΔtΔΦ=0.6822.11×10−5≈3.09×10−5V
Final Answer:
The magnitude of the average induced emf is approximately 3.09 × 10⁻⁵ V, or 30.9 µV.
Explanation:
- Faraday’s Law tells us that a changing magnetic flux induces an emf in the loop.
- The loop’s area changes as it shrinks, causing the magnetic flux to change, which leads to the induction of emf.
- The rate at which the magnetic flux changes over time gives us the average induced emf.
