VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end:Points): A rectangular wire loop with radius 10 cm and resistance R= 1 © BERL © 0)» agai field perpendicular to the plane of the loop steadily from 0 to 1 T within a time of 10 ms. Draw the wire loop and indicat: EL Determine the induced current around the loop.
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give answer in 2 step with explanation at the end of each step and final answer at the end:
Points): A rectangular wire loop with radius 10 cm and resistance R= 1 ©
BERL © 0)» agai field perpendicular to the plane of the loop
steadily from 0 to 1 T within a time of 10 ms. Draw the wire loop and indicat:
EL Determine the induced current around the loop.
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Asked by: Rahul
Created at: 2025-12-09 02:05:23
Asked by: Rahul
Created at: 2025-12-09 02:05:23
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## Step 1: Calculate the Induced EMF
**Given:**
- Radius of loop, \( r = 10\,\text{cm} = .1\,\text{m} \)
- Resistance, \( R = 1\,\Omega \)
- Magnetic field changes from \( \) to \( 1\,\text{T} \) in \( 10\,\text{ms} = .01\,\text{s} \)
- Magnetic field is **perpendicular** to the loop
**Induced EMF (\( \mathcal{E} \))** is given by Faraday's Law:
\[
\mathcal{E} = -\frac{d\Phi_B}{dt}
\]
where \( \Phi_B = B \cdot A \) and \( A = \pi r^2 \).
\[
A = \pi (.1)^2 = .0314\,\text{m}^2
\]
Change in magnetic field (\( \Delta B \)) = \( 1 - = 1\,\text{T} \)
So,
\[
\mathcal{E} = -\frac{\Delta B \cdot A}{\Delta t} = -\frac{1 \times .0314}{.01} = -3.14\,\text{V}
\]
**Explanation:**
We used Faraday's Law to find the emf induced in the loop due to the changing magnetic field.
---
## Step 2: Find the Induced Current and Its Direction
**Induced Current (\( I \))**:
\[
I = \frac{|\mathcal{E}|}{R} = \frac{3.14}{1} = 3.14\,\text{A}
\]
**Direction:**
- The magnetic field is **increasing into the page**.
- By **Lenz's Law**, the induced current will produce a magnetic field **opposing** the increase.
- To oppose an increase "into the page," the induced current must create a field **out of the page**.
- By the **right-hand rule**, this means the current flows **counterclockwise** when viewed from above.
**Explanation:**
We divided the emf by the resistance to get the induced current and used Lenz's Law to determine the direction.
---
## **Final Answers**
1. **Induced Current:** \( 3.14\,\text{A} \)
2. **Direction:** Counterclockwise around the loop (when viewed from above).
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