Question:
A long solenoid has \(500\) turns per meter and carries a current of \(5\,\text{A}\). Magnetic field inside the solenoid is: (\(\mu_0 = 4\pi \times 10^{-7}\))
A long solenoid has \(500\) turns per meter and carries a current of \(5\,\text{A}\). Magnetic field inside the solenoid is: (\(\mu_0 = 4\pi \times 10^{-7}\))
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Be careful with the difference between $n$ and $N$ in solenoid equations:
- $n$ is the turn density (turns per unit length).
- $N$ is the total number of turns.
If the total turns and length are given, you must calculate $n = \frac{N}{L}$ first.
- $n$ is the turn density (turns per unit length).
- $N$ is the total number of turns.
If the total turns and length are given, you must calculate $n = \frac{N}{L}$ first.
Updated On: May 21, 2026
- $2\pi \times 10^{-3}\text{ T}$
- $\pi \times 10^{-3}\text{ T}$
- $4\pi \times 10^{-3}\text{ T}$
- $10^{-3}\text{ T}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The question asks to compute the magnetic field intensity ($B$) inside a long solenoid of a given turn density carrying a specified electric current.
Step 2: Key Formula or Approach:
The magnetic field inside an ideal, long, air-core solenoid is uniform and given by Ampere's Law:
\[ B = \mu_0 \cdot n \cdot I \] where $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7}\text{ T m A}^{-1}$), $n$ is the number of turns per unit length (turns/meter), and $I$ is the current.
Step 3: Detailed Explanation:
• Let us write down the parameters provided:
Turn density, $n = 500\text{ turns/m}$
Current, $I = 5\text{ A}$
Permeability constant, $\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$
• We substitute these values into the solenoid magnetic field formula:
\[ B = (4\pi \times 10^{-7}) \cdot 500 \cdot 5 \]
• Grouping the numbers to simplify the arithmetic:
\[ B = 4\pi \times 10^{-7} \cdot 2500 \] \[ B = 10000\pi \times 10^{-7} \]
• Expressing $10000$ in scientific notation as $10^4$:
\[ B = \pi \times 10^4 \times 10^{-7} \] \[ B = \pi \times 10^{-3}\text{ T} \]
• This uniform magnetic field is directed along the axis of the solenoid, and its direction is determined by the Right-Hand Grip Rule.
Step 4: Final Answer:
The magnetic field inside the solenoid is $\pi \times 10^{-3}\text{ T}$.
The question asks to compute the magnetic field intensity ($B$) inside a long solenoid of a given turn density carrying a specified electric current.
Step 2: Key Formula or Approach:
The magnetic field inside an ideal, long, air-core solenoid is uniform and given by Ampere's Law:
\[ B = \mu_0 \cdot n \cdot I \] where $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7}\text{ T m A}^{-1}$), $n$ is the number of turns per unit length (turns/meter), and $I$ is the current.
Step 3: Detailed Explanation:
• Let us write down the parameters provided:
Turn density, $n = 500\text{ turns/m}$
Current, $I = 5\text{ A}$
Permeability constant, $\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$
• We substitute these values into the solenoid magnetic field formula:
\[ B = (4\pi \times 10^{-7}) \cdot 500 \cdot 5 \]
• Grouping the numbers to simplify the arithmetic:
\[ B = 4\pi \times 10^{-7} \cdot 2500 \] \[ B = 10000\pi \times 10^{-7} \]
• Expressing $10000$ in scientific notation as $10^4$:
\[ B = \pi \times 10^4 \times 10^{-7} \] \[ B = \pi \times 10^{-3}\text{ T} \]
• This uniform magnetic field is directed along the axis of the solenoid, and its direction is determined by the Right-Hand Grip Rule.
Step 4: Final Answer:
The magnetic field inside the solenoid is $\pi \times 10^{-3}\text{ T}$.
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