A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant a

Q: Why does a rotating charge create a magnetic field?

Moving charges constitute an electric current. According to electromagnetism, any electric current generates a magnetic field around it. Since the charge is moving in a circle, it acts like a circular current loop.

Q: How do we find the time period $T$?

The time period $T$ is the time taken for one complete rotation. It is related to the angular speed $\omega$ by the standard kinematic formula $T = \frac{2\pi}{\omega}$.

Q: Why do we use the formula for a steady current loop?

Because the charge is uniformly distributed along the ring and it is rotating at a constant angular speed, the amount of charge passing any point per second is constant. This perfectly mimics a steady, continuous direct current (DC).

Q: What if the charge was not uniformly distributed?

If the charge were clumped in one spot, the current would not be steady. It would be a pulsed current, and the magnetic field at the center would fluctuate over time rather than being constant.

Q: Why did we convert the radius from cm to m?

To maintain consistency in SI units. The magnetic field is given in Tesla, and the permeability constant $\mu_0$ is in SI units. Therefore, all distances must be in meters.

Question

A thin ring of $10	ext{ cm}$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40\pi	ext{ rad s}^{-1}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 	imes 10^{-9}	ext{ T}$, then the charge carried by the ring is close to ($\mu_0 = 4\pi 	imes 10^{-7}	ext{ N/A}^2$).

Options:

(A) $2 	imes 10^{-6}	ext{ C}$

(B) $3 	imes 10^{-5}	ext{ C}$

(C) $4 	imes 10^{-5}	ext{ C}$

(D) $7 	imes 10^{-6}	ext{ C}$

Sigma Percentile · Accepted Answer

### The Magic of a Rotating Charge

Imagine a thin ring suspended in space, carrying a uniformly distributed charge $Q$. Now, imagine this ring spinning about its central axis with a constant angular speed $\omega$. What happens?

In the world of electromagnetism, a moving charge is the fundamental building block of an electric current. Because the charge is uniformly distributed and rotating steadily, it perfectly mimics a continuous, steady current flowing in a circular loop. This is a beautiful conceptual bridge: kinematics creating electrodynamics.

### The Master Equation

To find the equivalent current $I$, we use the basic definition of current: the rate of flow of charge. In one complete rotation, the entire charge $Q$ passes any given point. The time taken for one rotation is the time period $T$, which is related to the angular speed by $T = \frac{2\pi}{\omega}$.

Therefore, the equivalent current is:
$$I = \frac{Q}{T} = \frac{Q\omega}{2\pi}$$

Now, we know from the Biot-Savart Law that a steady current $I$ in a circular loop of radius $R$ generates a magnetic field $B$ at its center, given by:
$$B = \frac{\mu_0 I}{2R}$$

Substituting our equivalent current into this formula, we get a direct relationship between the magnetic field and the rotating charge:
$$B = \frac{\mu_0}{2R} \left(\frac{Q\omega}{2\pi}ight) = \frac{\mu_0 Q \omega}{4\pi R}$$

### The Final Calculation

Our goal is to find the charge $Q$. Let's rearrange our master equation to isolate $Q$:
$$Q = \frac{4\pi R B}{\mu_0 \omega}$$

Now, we carefully substitute the given values. Remember to convert the radius from centimeters to meters to maintain SI unit consistency ($R = 0.1	ext{ m}$):
$$Q = \frac{4\pi (0.1) (3.8 	imes 10^{-9})}{(4\pi 	imes 10^{-7}) (40\pi)}$$

Notice how the $4\pi$ terms elegantly cancel out, simplifying our arithmetic:
$$Q = \frac{0.1 	imes 3.8 	imes 10^{-9}}{10^{-7} 	imes 40\pi}$$

$$Q = \frac{0.38 	imes 10^{-2}}{40\pi}$$

Evaluating this expression gives us:
$$Q \approx 3.022 	imes 10^{-5}	ext{ C}$$

This value is closest to **$3 	imes 10^{-5}	ext{ C}$**, which corresponds to option (b). It is a brilliant example of how mechanical rotation can be seamlessly translated into an electromagnetic phenomenon!

Select Answer:

Visualized Solution

Visualizing the Setup

Equivalent Current

Magnetic Field of a Loop

Substituting Current

Rearranging for Charge

Substituting Values

Simplification

Final Answer

The Sigma Insight: Biot-Savart Law

The Magic of a Rotating Charge

The Master Equation

The Final Calculation

Similar Questions

A battery is connected between two points $A$ and $B$ on the circumference of a uniform conducting ring of radius $r$ and resistance $R$ . One of the arcs $A B$ of the ring subtends an angle $θ$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is [Take, $μ_{0} = 4 π \times 1 0^{- 7} NA^{- 2}$ ]

The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is $54 μ T$ . What will be its value at the centre of the loop?

Two concentric coils each of radius equal to $2 π$ cm are placed at right angles to each other. 3 A and 4 A are the currents flowing in each coil respectively. The magnetic induction in $Wb/m^{2}$ at the centre of the coils will be ( $μ_{0} = 4 π \times 1 0^{- 7} Wb/Am$ )

A very long wire ABDMNDC is shown in figure carrying current $I$ . $A B$ and $B C$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $D M N D$ of radius $R$ . $A B$ , $B C$ parts are tangential to circular turn at $N$ and $D$ . Magnetic field at the centre of circle is

A current loop, having two circular arcs joined by two radial lines as shown in the figure. It carries a current of $10 A$ . The magnetic field at point $O$ will be close to

Find the magnetic field at point $P$ due to a straight line segment $A B$ of length $6 cm$ carrying a current of $5 A$ (See figure). (Take, $μ_{0} = 4 π \times 1 0^{- 7} N-A^{- 2}$ )

A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $I$ passes through the coil, the magnetic field at the centre is

Magnetic fields at two points on the axis of a circular coil at a distance of $0.05 m$ and $0.2 m$ from the centre are in the ratio $8 : 1$ . The radius of coil is