A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of

Q: Why do we use $\mu$ instead of $\mu_0$ in the formula?

Because the solenoid is not empty! It has a core with its own magnetic properties. $\mu_0$ is only used for a vacuum or air. For any other medium, we must use its absolute permeability $\mu = \mu_r \mu_0$.

Q: What does 'turns per metre' mean? Do I need to divide by length?

'Turns per metre' is already the turn density $n$ (where $n = N/L$). You don't need to divide by length again; that step is already done for you!

Q: Is magnetic flux density the same as the magnetic field?

Yes, in this context, magnetic flux density (denoted by $B$) is just another formal name for the magnetic field vector $\vec{B}$.

Q: Why did the powers of 10 disappear in the final calculation?

It's a beautiful mathematical coincidence designed into this problem! The product of the integers gave exactly $10^7$, which perfectly canceled out the $10^{-7}$ from the $\mu_0$ constant.

Q: What would happen if the current was reversed?

The magnitude of the magnetic field would remain exactly the same ($\pi \text{ T}$), but its direction would reverse, pointing in the opposite way along the central axis of the solenoid.

Question

A solenoid of $1000$ turns per metre has a core with relative permeability $500$. Insulated windings of the solenoid carry an electric current of $5	ext{ A}$. The magnetic flux density produced by the solenoid is
(Permeability of free space $= 4\pi 	imes 10^{-7}	ext{ H/m}$)

Options:

(A) $\pi	ext{ T}$

(B) $2 	imes 10^{-3} \pi	ext{ T}$

(C) $\frac{\pi}{5}	ext{ T}$

(D) $10^{-4} \pi	ext{ T}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a long cylindrical core made of a special magnetic material. We wrap an insulated wire around it to form a solenoid. When we pass a current of $5	ext{ A}$ through this wire, a strong magnetic field is generated inside.

We are given several key parameters. The turn density of the solenoid is $n = 1000	ext{ turns/m}$. The relative permeability of the core material is $\mu_r = 500$. The current flowing through the windings is $I = 5	ext{ A}$. Our goal is to find the magnetic flux density, which is simply the magnetic field $\vec{B}$ inside the solenoid.

### The Master Equation

To find the magnetic field inside a long solenoid, we rely on a classic result derived from Ampere's Law. The magnetic field $B$ is given by the product of the medium's permeability $\mu$, the turn density $n$, and the current $I$:

$$B = \mu n I$$

However, the solenoid is not empty; it contains a core. The actual absolute permeability of this core $\mu$ is the product of its relative permeability $\mu_r$ and the permeability of free space $\mu_0$:

$$\mu = \mu_r \mu_0$$

By substituting this relationship into our initial equation, we obtain the master formula that connects all our given variables:

$$B = \mu_r \mu_0 n I$$

### Final Calculation

Now, let's substitute the given values into our master equation. We know that $\mu_0 = 4\pi 	imes 10^{-7}	ext{ H/m}$.

$$B = (500) 	imes (4\pi 	imes 10^{-7}) 	imes (1000) 	imes (5)$$

Instead of rushing to a calculator, let's group the integers and the powers of $10$ to simplify the expression elegantly. First, multiply the scalar numbers:

$$500 	imes 5 	imes 1000 	imes 4 = 2500 	imes 4000 = 10,000,000 = 10^7$$

Now, bring back the $\pi$ and the $10^{-7}$:

$$B = 10^7 	imes \pi 	imes 10^{-7}$$

Notice how beautifully the powers of ten cancel out! $10^7 	imes 10^{-7} = 10^0 = 1$. We are left with a remarkably clean final answer:

**$$B = \pi 	ext{ T}$$**

This perfectly matches option (a). This problem beautifully illustrates how a ferromagnetic core can amplify a magnetic field by a factor of its relative permeability!

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Visualized Solution

$Visualizing the Solenoid$

$Magnetic Field Formula$

$Absolute Permeability$

$Combined Equation$

$Substituting Values$

$Simplifying the Expression$

$Final Answer$

$Conceptual Takeaway$

The Sigma Insight: Ampere's Circuital Law

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Solenoid

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Absolute Permeability

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Combined Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conceptual Takeaway