A wire A, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm

Q: Why do we have to convert the angle to radians? Can't we just use degrees?

The standard formula $B = \frac{\mu_0 I}{4\pi R} \theta$ strictly requires $\theta$ to be in radians because it is derived using the arc length formula $l = R\theta$. If you want to use degrees, you must use the modified formula $B = \frac{\mu_0 I}{2R} \times \frac{\theta}{360^\circ}$.

Q: Why did we ignore the straight dashed lines in the diagram?

The straight lines in the diagram are merely geometric indicators to show you the angle of the missing sector. The current only flows through the solid circular arc, so only the arc contributes to the magnetic field at the center.

Q: Does the direction of the current (clockwise vs counter-clockwise) matter for this ratio?

The direction of the current determines the vector direction of the magnetic field (into or out of the page). However, since the question asks for the ratio of their magnitudes, the direction does not affect the final numerical calculation.

Q: What if the two arcs were placed concentrically in the same plane?

If they were concentric and you needed to find the *net* magnetic field, you would have to consider their directions. If the currents flow in the same sense, you add their magnitudes. If they flow in opposite senses, you subtract them.

Q: I got $12/10$ but the options don't have it. What did I do wrong?

You didn't do anything wrong! You just need to simplify the fraction. Both $12$ and $10$ are divisible by $2$, which simplifies the ratio to $6/5$ or $6:5$.

Question

A wire $A$, bent in the shape of an arc of a circle, carrying a current of $2	ext{ A}$ and having radius $2	ext{ cm}$ and another wire $B$, also bent in the shape of arc of a circle, carrying a current of $3	ext{ A}$ and having radius of $4	ext{ cm}$, are placed as shown in the figure. The ratio of the magnetic fields due to the wires $A$ and $B$ at the common centre $O$ is

Options:

(A) $4 : 6$

(B) $6 : 4$

(C) $6 : 5$

(D) $2 : 5$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at two distinct circular arcs, let's call them Wire $A$ and Wire $B$. Each of these wires carries a different amount of current and has a different radius. Our objective is to determine the ratio of the magnetic fields they produce at their respective centers, which we will call point $O$.

### The Master Equation

To tackle this problem, we need to recall the fundamental formula for the magnetic field at the center of a circular arc. The magnetic field $B$ depends on the current $I$, the radius $R$, and the angle $	heta$ subtended by the arc at the center. The formula is given by:

$$B = \frac{\mu_0 I}{4\pi R} 	heta$$

**Crucial Note:** In this formula, the angle $	heta$ must strictly be in radians.

### Extracting Parameters for Wire A

Let's break down the parameters for Wire $A$. The problem states that the current $I_A$ is $2	ext{ A}$ and the radius $R_A$ is $2	ext{ cm}$, which is $2 	imes 10^{-2}	ext{ m}$.

Looking at the diagram for Wire $A$, we see a full circle with a $90^\circ$ sector missing. Therefore, the angle subtended by the arc is:

$$	heta_A = 360^\circ - 90^\circ = 270^\circ$$

Converting this to radians, we get $	heta_A = \frac{3\pi}{2}	ext{ rad}$.

### Extracting Parameters for Wire B

Now, let's do the same for Wire $B$. The current $I_B$ is $3	ext{ A}$ and the radius $R_B$ is $4	ext{ cm}$, or $4 	imes 10^{-2}	ext{ m}$.

For Wire $B$, the diagram shows a $60^\circ$ sector missing from the full circle. Thus, the angle is:

$$	heta_B = 360^\circ - 60^\circ = 300^\circ$$

Converting this to radians gives us $	heta_B = \frac{5\pi}{3}	ext{ rad}$.

### Setting Up the Ratio

We are asked to find the ratio of the magnetic fields, $\frac{B_A}{B_B}$. Let's set up the equation using our master formula:

$$\frac{B_A}{B_B} = \frac{\frac{\mu_0 I_A}{4\pi R_A} 	heta_A}{\frac{\mu_0 I_B}{4\pi R_B} 	heta_B}$$

Notice how the constant terms $\frac{\mu_0}{4\pi}$ beautifully cancel out from the numerator and the denominator. This leaves us with a much simpler expression:

$$\frac{B_A}{B_B} = \frac{	heta_A I_A R_B}{	heta_B I_B R_A}$$

### Final Calculation

Now, we carefully substitute all our extracted values into this simplified ratio. Don't rush through this step; ensure every value is placed correctly.

$$\frac{B_A}{B_B} = \frac{\left(\frac{3\pi}{2}ight) 	imes 2 	imes 4 	imes 10^{-2}}{\left(\frac{5\pi}{3}ight) 	imes 3 	imes 2 	imes 10^{-2}}$$

We can immediately cancel out the $10^{-2}$ terms. Let's simplify the numerator and the denominator:

$$\frac{B_A}{B_B} = \frac{3\pi 	imes 4}{5\pi 	imes 2}$$

The $\pi$ terms also cancel out.

$$\frac{B_A}{B_B} = \frac{12}{10}$$

Finally, dividing both the numerator and the denominator by $2$, we get our elegant final answer:

**$$\frac{B_A}{B_B} = \frac{6}{5}$$**

So, the ratio of the magnetic fields $B_A : B_B$ is **$6 : 5$**.

Select Answer:

Visualized Solution

Analyzing the Setup

Magnetic Field of a Circular Arc

Parameters for Wire A

Parameters for Wire B

Setting up the Ratio

Substituting the Values

Calculating the Final Ratio

Further Exploration

The Sigma Insight: Biot-Savart Law

Analyzing the Setup

The Master Equation

Extracting Parameters for Wire A

Extracting Parameters for Wire B

Setting Up the Ratio

Final Calculation

Similar Questions

If in a circular coil $A$ of radius $R$ , current $I$ is flowing and in another coil $B$ of radius $2 R$ a current $2 I$ is flowing, then the ratio of the magnetic fields, $B_{A}$ and $B_{B}$ produced by them will be

There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductors is 1 : 1. The magnetic field at point P is

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

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

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 ( $μ_{0} = 4 π \times 1 0^{- 7} Wb/Am$ )

Select Answer:

Visualized Solution

Analyzing the Setup

Magnetic Field of a Circular Arc

Parameters for Wire A

Parameters for Wire B

Setting up the Ratio

Substituting the Values

Calculating the Final Ratio

Further Exploration

The Sigma Insight: Biot-Savart Law

Analyzing the Setup

The Master Equation

Extracting Parameters for Wire A

Extracting Parameters for Wire B

Setting Up the Ratio

Final Calculation

Similar Questions

There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductors is 1 : 1. The magnetic field at point P is

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

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