Question:
A circular coil, carrying current, has radius \( R \). The distance from the centre of the coil on the axis where the magnetic induction will be \( \frac{1}{27}\text{th} \) of its value at the centre of the coil is
A circular coil, carrying current, has radius \( R \). The distance from the centre of the coil on the axis where the magnetic induction will be \( \frac{1}{27}\text{th} \) of its value at the centre of the coil is
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To solve questions involving \( B(x) = \frac{1}{n} B_0 \) quickly, use the direct relation:
\[ \left( 1 + \frac{x^2}{R^2} \right) = n^{2/3} \]
For \( n = 27 \), we get \( n^{2/3} = (27)^{2/3} = 9 \). Thus, \( \frac{x^2}{R^2} = 9 - 1 = 8 \implies x = \sqrt{8}R = 2\sqrt{2}R \). This shortcut saves valuable calculation time.
\[ \left( 1 + \frac{x^2}{R^2} \right) = n^{2/3} \]
For \( n = 27 \), we get \( n^{2/3} = (27)^{2/3} = 9 \). Thus, \( \frac{x^2}{R^2} = 9 - 1 = 8 \implies x = \sqrt{8}R = 2\sqrt{2}R \). This shortcut saves valuable calculation time.
Updated On: May 25, 2026
- \( 2\sqrt{2} R \)
- \( 3\sqrt{2} R \)
- \( 3 R \)
- \( 2\sqrt{3} R \)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
We need to find the axial distance \( x \) from the center of a current-carrying circular coil of radius \( R \) where the magnetic induction is \( \frac{1}{27}\text{th} \) of the magnetic induction at its center.
Step 2: Key Formula or Approach:
The magnetic field \( B(x) \) at a distance \( x \) on the axis of a circular coil of radius \( R \) with \( N \) turns carrying current \( I \) is:
\[ B(x) = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} \]
The magnetic field \( B_0 \) at the center of the coil (\( x = 0 \)) is:
\[ B_0 = \frac{\mu_0 N I}{2R} \]
Step 3: Detailed Explanation:
According to the problem:
\[ B(x) = \frac{1}{27} B_0 \]
Substituting the expressions for \( B(x) \) and \( B_0 \):
\[ \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{27} \left( \frac{\mu_0 N I}{2R} \right) \]
Canceling the common term \( \frac{\mu_0 N I}{2} \) on both sides:
\[ \frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{27 R} \]
Multiply both sides by \( R \):
\[ \frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{27} \]
Taking the reciprocal of both sides:
\[ \left( \frac{R^2 + x^2}{R^2} \right)^{3/2} = 27 \]
Taking the cube root on both sides:
\[ \left( \frac{R^2 + x^2}{R^2} \right)^{1/2} = (27)^{1/3} = 3 \]
Squaring both sides:
\[ \frac{R^2 + x^2}{R^2} = 9 \]
\[ R^2 + x^2 = 9 R^2 \]
\[ x^2 = 8 R^2 \]
\[ x = \sqrt{8} R = 2\sqrt{2} R \]
Step 4: Final Answer:
The distance from the center of the coil is \( 2\sqrt{2} R \), which corresponds to option (A).
We need to find the axial distance \( x \) from the center of a current-carrying circular coil of radius \( R \) where the magnetic induction is \( \frac{1}{27}\text{th} \) of the magnetic induction at its center.
Step 2: Key Formula or Approach:
The magnetic field \( B(x) \) at a distance \( x \) on the axis of a circular coil of radius \( R \) with \( N \) turns carrying current \( I \) is:
\[ B(x) = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} \]
The magnetic field \( B_0 \) at the center of the coil (\( x = 0 \)) is:
\[ B_0 = \frac{\mu_0 N I}{2R} \]
Step 3: Detailed Explanation:
According to the problem:
\[ B(x) = \frac{1}{27} B_0 \]
Substituting the expressions for \( B(x) \) and \( B_0 \):
\[ \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{27} \left( \frac{\mu_0 N I}{2R} \right) \]
Canceling the common term \( \frac{\mu_0 N I}{2} \) on both sides:
\[ \frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{27 R} \]
Multiply both sides by \( R \):
\[ \frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{27} \]
Taking the reciprocal of both sides:
\[ \left( \frac{R^2 + x^2}{R^2} \right)^{3/2} = 27 \]
Taking the cube root on both sides:
\[ \left( \frac{R^2 + x^2}{R^2} \right)^{1/2} = (27)^{1/3} = 3 \]
Squaring both sides:
\[ \frac{R^2 + x^2}{R^2} = 9 \]
\[ R^2 + x^2 = 9 R^2 \]
\[ x^2 = 8 R^2 \]
\[ x = \sqrt{8} R = 2\sqrt{2} R \]
Step 4: Final Answer:
The distance from the center of the coil is \( 2\sqrt{2} R \), which corresponds to option (A).
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