Problem 1. The wire is moving to the left with a linear velocity of 10 m/s as shown. Determine the magnitude and direction of the emf induced in the wire. Include a brief explanation. 45° XV= 10 m/s 1-025 m B- 02 T, into the page XE X
Problem 1. The wire is moving to the left with a linear velocity of 10 m/s as shown. Determine the magnitude and direction of the emf induced in the wire. Include a brief explanation. 45° XV= 10 m/s 1-025 m B- 02 T, into the page XE X
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![**Problem 1:**
The wire is moving to the left with a linear velocity of \(10 \, \text{m/s}\) as shown. Determine the magnitude and direction of the emf induced in the wire. Include a brief explanation.
**Diagram Explanation:**
- The wire is oriented along a path at a \(45^\circ\) angle to the horizontal plane.
- It is moving with a velocity \(v = 10 \, \text{m/s}\) to the left.
- The length of the wire segment within the magnetic field is \(l = 0.25 \, \text{m}\).
- A uniform magnetic field \(B = 0.2 \, \text{T}\) is directed into the page, represented by crosses.
**Solution Approach:**
To find the induced emf (\(\epsilon\)), use the formula:
\[
\epsilon = B \cdot v \cdot l \cdot \sin{\theta}
\]
where:
- \(B\) is the magnetic field strength,
- \(v\) is the velocity of the wire,
- \(l\) is the length of the wire,
- \(\theta\) is the angle between the velocity and the magnetic field (since it is into the page, directly \(\theta = 90^\circ - 45^\circ = 45^\circ\)).
Plug in the given values to calculate the magnitude of the emf. The direction of the induced emf can be determined by using the right-hand rule.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf1d0bb8-6fc4-423e-9784-469bf31e55bd%2Fff8fbbce-4c61-40eb-b0fe-81ab4ed6dfde%2Fo8zgia_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 1:**
The wire is moving to the left with a linear velocity of \(10 \, \text{m/s}\) as shown. Determine the magnitude and direction of the emf induced in the wire. Include a brief explanation.
**Diagram Explanation:**
- The wire is oriented along a path at a \(45^\circ\) angle to the horizontal plane.
- It is moving with a velocity \(v = 10 \, \text{m/s}\) to the left.
- The length of the wire segment within the magnetic field is \(l = 0.25 \, \text{m}\).
- A uniform magnetic field \(B = 0.2 \, \text{T}\) is directed into the page, represented by crosses.
**Solution Approach:**
To find the induced emf (\(\epsilon\)), use the formula:
\[
\epsilon = B \cdot v \cdot l \cdot \sin{\theta}
\]
where:
- \(B\) is the magnetic field strength,
- \(v\) is the velocity of the wire,
- \(l\) is the length of the wire,
- \(\theta\) is the angle between the velocity and the magnetic field (since it is into the page, directly \(\theta = 90^\circ - 45^\circ = 45^\circ\)).
Plug in the given values to calculate the magnitude of the emf. The direction of the induced emf can be determined by using the right-hand rule.
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