Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from

Q: Why did the $r^2$ in the numerator cancel out when taking the ratio?

Because the coil is the same for both points, its radius $r$ and the current $I$ are constant. When you divide $B_1$ by $B_2$, any term that is identical in both the numerator and denominator will cancel out.

Q: How do I handle the $3/2$ power in the equation?

The $3/2$ power means 'take the cube, then take the square root' (or vice versa). By recognizing that the other side of the equation is 8 (which is $2^3$), you can equate the cubed parts, leaving you with a $1/2$ power (square root) equal to 2. Squaring both sides then removes the root.

Q: What if the ratio wasn't a perfect cube like 8?

In JEE problems, they usually are! If it wasn't, you would have to take the $2/3$ power of the ratio using logarithms or approximations, but that is extremely rare in this context.

Q: Why did we get $d_2$ in the numerator and $d_1$ in the denominator after taking the ratio?

Because the distance term $(d^2 + r^2)^{3/2}$ is in the denominator of the magnetic field formula. When you divide $B_1$ by $B_2$, the denominator of $B_2$ flips and goes to the numerator.

Q: Can the radius be negative?

Mathematically, $r^2 = 0.01$ gives $r = \pm 0.1$. However, in physics, the radius of a physical object like a coil must be a positive quantity, so we only take the positive root.

Question

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

Options:

(A) $0.2	ext{ m}$

(B) $0.1	ext{ m}$

(C) $0.15	ext{ m}$

(D) $1.0	ext{ m}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing on the axis of a circular coil carrying a steady current. As you move away from the center, the magnetic field weakens. In this problem, we are given two specific points on this axis, $P_1$ and $P_2$, located at distances $d_1 = 0.05	ext{ m}$ and $d_2 = 0.2	ext{ m}$ from the center.

We are told that the magnetic field at the closer point is exactly 8 times stronger than at the farther point. Our mission? To find the radius $r$ of this coil.

### The Master Equation
To unlock this, we need the Biot-Savart law's application for the magnetic field on the axis of a circular loop. The formula is:
$$B = \frac{\mu_0 I r^2}{2(d^2 + r^2)^{3/2}}$$

Notice how the field depends on the distance $d$. It's not a simple inverse square law because of the radius $r$ in the denominator.

### Setting Up the Ratio
Let's take the ratio of the magnetic fields at the two points. The beautiful thing about ratios in physics is that all the messy constants—like $\mu_0$, the current $I$, and the numerator $r^2$—cancel out perfectly!

$$\frac{B_1}{B_2} = \frac{(d_2^2 + r^2)^{3/2}}{(d_1^2 + r^2)^{3/2}}$$

We know this ratio is equal to 8. So, we have:
$$8 = \left(\frac{d_2^2 + r^2}{d_1^2 + r^2}ight)^{3/2}$$

### The Algebraic Magic
This equation might look intimidating with that $3/2$ power. But here is a pro-tip: always look for perfect cubes or squares! We can write 8 as $2^3$.

$$2^3 = \left(\frac{d_2^2 + r^2}{d_1^2 + r^2}ight)^{3/2}$$

By taking the cube root of both sides, the power of 3 vanishes:
$$2 = \left(\frac{d_2^2 + r^2}{d_1^2 + r^2}ight)^{1/2}$$

Now, simply square both sides to eliminate the square root:
$$4 = \frac{d_2^2 + r^2}{d_1^2 + r^2}$$

### Final Calculation
We have reduced a complex physics problem into a basic linear equation in $r^2$. Let's cross-multiply:
$$4(d_1^2 + r^2) = d_2^2 + r^2$$
$$4d_1^2 + 4r^2 = d_2^2 + r^2$$
$$3r^2 = d_2^2 - 4d_1^2$$

Now, it's time to plug in our given values: $d_1 = 0.05	ext{ m}$ and $d_2 = 0.2	ext{ m}$.
$$3r^2 = (0.2)^2 - 4(0.05)^2$$
$$3r^2 = 0.04 - 4(0.0025)$$
$$3r^2 = 0.04 - 0.01$$
$$3r^2 = 0.03$$

Dividing by 3, we get:
$$r^2 = 0.01$$

Taking the square root, we find the radius of the coil:
**$$r = 0.1	ext{ m}$$**

And there we have it! The radius of the coil is $0.1	ext{ m}$. This problem is a classic example of how a scary-looking equation can collapse into a simple calculation if you handle the algebra with patience.

Animated Solution for Physics - Magnetic Effects of Current: 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

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Visualized Solution

$T h e S e t u p$

$M a g n e t i c F i e l d o n A x i s$

$R a t i o o f M a g n e t i c F i e l d s$

$E q u a t in g t h e R a t i o$

$S im pl i f y in g t h e P o w er$

$C r oss M u l t i pl i c a t i o n$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t in g R a d i u s$

$C o n c l u s i o n$

The Sigma Insight: Biot-Savart Law

Analyzing the Setup Imagine you are standing on the axis of a circular coil carrying a steady current

The Master Equation To unlock this, we need the Biot-Savart law's application for the magnetic field on the axis of a circular loop

Setting Up the Ratio Let's take the ratio of the magnetic fields at the two points

The Algebraic Magic This equation might look intimidating with that $3/2$ power

Final Calculation We have reduced a complex physics problem into a basic linear equation in $r^{2}$

Similar Questions

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Visualized Solution

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

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

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

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

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

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

.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; } Calculating Radius

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