Arrange the following in asending order of magnetic induction at a point on axis of circular loop of radius (r) at a distance (x) from its centre (for the same currents in the loops)
a) `r=R,x=sqrt3R` b) `r=2R,x=sqrt5R`
c) `r=R/2,x=2sqrt2R`

Options

A. a, c, b
B. a, b, c
C. b, c, a
D. c, a, b

Answer

To solve the problem of arranging the magnetic induction at a point on the axis of a circular loop in ascending order, we can follow these steps:

### Step 1: Understand the Formula for Magnetic Induction
The magnetic induction \( B \) at a point on the axis of a circular loop of radius \( r \) at a distance \( x \) from its center is given by the formula:
\[ B = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current flowing through the loop,
- \( r \) is the radius of the loop,
- \( x \) is the distance from the center of the loop.

### Step 2: Calculate Magnetic Induction for Each Case
We will calculate \( B \) for each of the three cases provided.

#### Case A: \( r = R, x = \sqrt{3}R \)
Using the formula:
\[ B_A = \frac{\mu_0 I R^2}{2(R^2 + (\sqrt{3}R)^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(8R^3)} = \frac{\mu_0 I}{16R} \]

#### Case B: \( r = 2R, x = \sqrt{5}R \)
Using the formula:
\[ B_B = \frac{\mu_0 I (2R)^2}{2((2R)^2 + (\sqrt{5}R)^2)^{3/2}} = \frac{\mu_0 I (4R^2)}{2(4R^2 + 5R^2)^{3/2}} = \frac{\mu_0 I (4R^2)}{2(9R^2)^{3/2}} = \frac{\mu_0 I (4R^2)}{2(27R^3)} = \frac{2\mu_0 I}{27R} \]

#### Case C: \( r = \frac{R}{2}, x = 2\sqrt{2}R \)
Using the formula:
\[ B_C = \frac{\mu_0 I \left(\frac{R}{2}\right)^2}{2\left(\left(\frac{R}{2}\right)^2 + (2\sqrt{2}R)^2\right)^{3/2}} = \frac{\mu_0 I \left(\frac{R^2}{4}\right)}{2\left(\frac{R^2}{4} + 8R^2\right)^{3/2}} = \frac{\mu_0 I \left(\frac{R^2}{4}\right)}{2\left(\frac{33R^2}{4}\right)^{3/2}} = \frac{\mu_0 I \left(\frac{R^2}{4}\right)}{2\left(\frac{33^{3/2}R^3}{8}\right)} = \frac{4\mu_0 I}{33^{3/2}R} \]

### Step 3: Compare the Values
Now we have:
- \( B_A = \frac{\mu_0 I}{16R} \)
- \( B_B = \frac{2\mu_0 I}{27R} \)
- \( B_C = \frac{4\mu_0 I}{33^{3/2}R} \)

To compare these values, we can ignore \( \mu_0 I/R \) since they are common in all cases.

1. Compare \( \frac{1}{16} \), \( \frac{2}{27} \), and \( \frac{4}{33^{3/2}} \).
2. Calculate numerical approximations:
- \( \frac{1}{16} = 0.0625 \)
- \( \frac{2}{27} \approx 0.0741 \)
- \( \frac{4}{33^{3/2}} \approx 0.0697 \)

### Step 4: Arrange in Ascending Order
From the calculations:
- \( B_A < B_C < B_B \)

Thus, the ascending order of magnetic induction is:
\[ B_C < B_A < B_B \]

### Final Answer
The correct ascending order is:
**Case C < Case A < Case B**