Question:
A closely wound solenoid of length 1 m has 5 layers of 500 turns each. If the magnitude of the magnetic field inside the solenoid near its centre is 4.4 mT, find the current carried.
A closely wound solenoid of length 1 m has 5 layers of 500 turns each. If the magnitude of the magnetic field inside the solenoid near its centre is 4.4 mT, find the current carried.
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Magnetic field inside solenoid: \(B = \mu_0 n I\), total turns per unit length includes all layers.
Updated On: Jul 18, 2026
- 1.4 A
- 1.5 A
- 1.6 A
- 1.8 A
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The Correct Option is A
Solution and Explanation
Step 1: Recall magnetic field formula for solenoid.
\[ B = \mu_0 n I \]
where \(n\) is number of turns per unit length, \(I\) is current, \(B\) is magnetic field, \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\).
Step 2: Compute total number of turns.
5 layers of 500 turns each: \(N = 5 \cdot 500 = 2500\) turns.
Step 3: Compute turns per unit length.
\[ n = \frac{N}{L} = \frac{2500}{1} = 2500 \, \text{turns/m} \]
Step 4: Solve for current.
\[ I = \frac{B}{\mu_0 n} = \frac{4.4 \times 10^{-3}}{4\pi \times 10^{-7} \cdot 2500} \]
Step 5: Simplify.
\[ I \approx 1.4 \, \text{A} \]
Step 6: Final conclusion.
Hence, the current carried by the solenoid is:
\[ \boxed{1.4 \, \text{A}} \]
\[ B = \mu_0 n I \]
where \(n\) is number of turns per unit length, \(I\) is current, \(B\) is magnetic field, \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\).
Step 2: Compute total number of turns.
5 layers of 500 turns each: \(N = 5 \cdot 500 = 2500\) turns.
Step 3: Compute turns per unit length.
\[ n = \frac{N}{L} = \frac{2500}{1} = 2500 \, \text{turns/m} \]
Step 4: Solve for current.
\[ I = \frac{B}{\mu_0 n} = \frac{4.4 \times 10^{-3}}{4\pi \times 10^{-7} \cdot 2500} \]
Step 5: Simplify.
\[ I \approx 1.4 \, \text{A} \]
Step 6: Final conclusion.
Hence, the current carried by the solenoid is:
\[ \boxed{1.4 \, \text{A}} \]
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