A long circular tube of length 10 m and radius 0.3 m carries a current I along its curved surface as

Q: Why can we treat the tube as a solenoid?

A long circular tube with current flowing along its curved surface is mathematically equivalent to a tightly wound long solenoid. The current per unit length $I/L$ acts just like $ni$ in a solenoid, producing a uniform magnetic field inside.

Q: Why is the magnetic field outside the tube zero?

Just like an ideal long solenoid, the magnetic field lines outside cancel each other out, resulting in a net zero magnetic field outside the tube.

Q: Why do we multiply by $\pi r^2$ twice?

We multiply by the area $\pi r^2$ once to find the magnetic flux ($\phi = B \cdot A$) and a second time to find the magnetic moment ($M = i \cdot A$).

Q: What does the negative sign in Faraday's Law signify here?

The negative sign represents Lenz's Law, indicating that the induced current will flow in a direction to oppose the change in magnetic flux. However, when we differentiate the cosine function, we get another negative sign, making the final expression positive.

Q: Why do we approximate $0.6 \pi^2$ to 6?

The value of $\pi^2$ is approximately 9.87. Multiplying this by 0.6 gives about 5.92. Since the question asks for an integer answer, we round it to the nearest integer, which is 6.

Question

A long circular tube of length 10 m and radius 0.3 m carries a current $I$ along its curved surface as shown. A wire-loop of resistance $0.005\ \Omega$ and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $I = I_0 \cos 300 t$ where $I_0$ is constant. If the magnetic moment of the loop is $N \mu_0 I_0 \sin (300 t)$, then $N$ is

Sigma Percentile · Accepted Answer

### The Setup: A Tube in Disguise

Imagine you are looking at this long circular tube. It might seem like just a simple cylinder, but physically, it is hiding a secret. Because the current $I$ flows along its curved surface, it behaves exactly like a tightly wound **long solenoid**.

This is a crucial realization. For a solenoid, the current per unit length is the total current divided by the length, which gives us $n = \frac{I}{L}$.

### The Magnetic Field and Flux

Because it acts like a long solenoid, it creates a uniform magnetic field inside its core. The formula for this magnetic field is given by Ampere's Law as $B = \mu_0 n$.

Substituting our current per unit length, we get $B = \frac{\mu_0 I}{L}$.

Now, let's focus on the small wire loop placed inside this tube. The magnetic field lines pass directly through it. The magnetic flux $\phi$ linked with this loop is simply the magnetic field multiplied by the loop's area.

Therefore, $\phi = B \cdot A = \left( \frac{\mu_0 I}{L} ight) (\pi r^2)$.

### Faraday's Law and Induced Current

Here is where the dynamics kick in. The main current $I$ is not constant; it is changing with time! This means the magnetic flux through the small loop is also changing.

According to **Faraday's Law of Induction**, a changing magnetic flux induces an Electromotive Force (EMF) in the loop. We find this by differentiating the flux with respect to time: $e = - \frac{d\phi}{dt} = - \left( \frac{\mu_0 \pi r^2}{L} ight) \frac{dI}{dt}$.

This induced EMF acts like a battery, driving an induced current $i$ in the small loop. By Ohm's law, this current is the EMF divided by the loop's resistance $R$.

So, $i = \frac{e}{R} = - \left( \frac{\mu_0 \pi r^2}{L R} ight) \frac{dI}{dt}$.

### The Magnetic Moment

A current-carrying loop is essentially a tiny magnet, and its strength is measured by its **magnetic moment** $M$. The magnetic moment is the product of the induced current $i$ and the loop's area $\pi r^2$.

Substituting our expression for the induced current, we get $M = i (\pi r^2) = - \left( \frac{\mu_0 \pi^2 r^4}{L R} ight) \frac{dI}{dt}$.

We are given that the main current varies as $I = I_0 \cos(300 t)$. To find the rate of change of current, we differentiate this with respect to time.

The derivative of cosine is negative sine, yielding $\frac{dI}{dt} = -300 I_0 \sin(300 t)$.

### The Final Calculation

Let's plug this derivative back into our magnetic moment equation. The negative signs beautifully cancel out, leaving us with $M = \left( \frac{300 \pi^2 r^4}{L R} ight) \mu_0 I_0 \sin(300 t)$.

The problem states that the magnetic moment is $N \mu_0 I_0 \sin(300 t)$. By comparing the two expressions, we can isolate $N$:

$N = \frac{300 \pi^2 r^4}{L R}$.

Now, it is just a matter of substituting the given numerical values: $r = 0.1	ext{ m}$, $L = 10	ext{ m}$, and $R = 0.005\ \Omega$.

Calculating this gives $N = \frac{300 \pi^2 (0.1)^4}{10 	imes 0.005} = 0.6 \pi^2$.

Since $\pi^2 \approx 9.87$, we find $N \approx 5.92$. Rounding to the nearest integer, we get our final answer: **6**.

Enter Numerical Value:

Visualized Solution

$n = \frac{I}{L}$

$B = \frac{μ _{0} I}{L}$

$ϕ = B \cdot A$

$e = - \frac{d ϕ}{d t}$

$i = \frac{e}{R}$

$M = i \cdot A$

$\frac{d I}{d t}$

$M (t)$

$N \approx 6$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

The Setup

The Magnetic Field and Flux

Faraday's Law and Induced Current

The Magnetic Moment

The 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; } n=LI​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B=Lμ0​I​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ϕ=B⋅A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e=−dtdϕ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=Re​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M=i⋅A

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M(t)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N≈6