Consider a cylindrical solenoid with diameter = d, length = λ, and number of turns of wire = n. If the wire carries current i, then what is magnetic flux through the circular cross-sectional area of the soleno d = 6.09 cm; X 19.40 cm; n = 216; i = 19.8 A.
Consider a cylindrical solenoid with diameter = d, length = λ, and number of turns of wire = n. If the wire carries current i, then what is magnetic flux through the circular cross-sectional area of the soleno d = 6.09 cm; X 19.40 cm; n = 216; i = 19.8 A.
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![**Problem Statement:**
Consider a cylindrical solenoid with:
- Diameter, \( d = 6.09 \, \text{cm} \)
- Length, \( \lambda = 19.40 \, \text{cm} \)
- Number of turns of wire, \( n = 216 \)
If the wire carries a current \( i = 19.8 \, \text{A} \), then what is the magnetic flux through the circular cross-sectional area of the solenoid?
**Solution:**
To find the magnetic flux through the solenoid, we need to use the formula for the magnetic flux through a solenoid:
\[ \Phi = B \cdot A \]
Where:
- \( B \) is the magnetic field inside the solenoid \((B = \mu_0 \cdot \frac{n}{\lambda} \cdot i)\)
- \( A \) is the cross-sectional area of the solenoid \((A = \pi \cdot (r)^2)\), where \( r \) is the radius of the solenoid.
Substitute the given values into the formula to calculate the magnetic flux.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F007d2f28-0b15-4cef-98f8-cbb273656312%2F91396026-4daf-49c2-914c-bbab5b668cca%2Foq4w3m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Consider a cylindrical solenoid with:
- Diameter, \( d = 6.09 \, \text{cm} \)
- Length, \( \lambda = 19.40 \, \text{cm} \)
- Number of turns of wire, \( n = 216 \)
If the wire carries a current \( i = 19.8 \, \text{A} \), then what is the magnetic flux through the circular cross-sectional area of the solenoid?
**Solution:**
To find the magnetic flux through the solenoid, we need to use the formula for the magnetic flux through a solenoid:
\[ \Phi = B \cdot A \]
Where:
- \( B \) is the magnetic field inside the solenoid \((B = \mu_0 \cdot \frac{n}{\lambda} \cdot i)\)
- \( A \) is the cross-sectional area of the solenoid \((A = \pi \cdot (r)^2)\), where \( r \) is the radius of the solenoid.
Substitute the given values into the formula to calculate the magnetic flux.
Expert Solution
Step 1: Given
Diameter (d) = 6.09 cm
Length (λ)= 19.40 cm
Number of loop (n) = 216
Current ( i ) = 19.8 A
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