Question:
The magnetic field due to a current carrying circular loop of radius \(6\ \text{cm}\) at a point on the axis at a distance of \(8\ \text{cm}\) from its centre is \(27\ \mu\text{T}\). The magnetic field at the centre of the current carrying loop is
The magnetic field due to a current carrying circular loop of radius \(6\ \text{cm}\) at a point on the axis at a distance of \(8\ \text{cm}\) from its centre is \(27\ \mu\text{T}\). The magnetic field at the centre of the current carrying loop is
Show Hint
For a circular current loop,
\[
B_{\text{centre}}=\frac{\mu_0 I}{2R},
\]
and at a point on the axis,
\[
B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}.
\]
Taking the ratio of these two formulas often eliminates the current \(I\) and simplifies the calculation.
Updated On: Jun 26, 2026
- \(75\ \mu\text{T}\)
- \(125\ \mu\text{T}\)
- \(150\ \mu\text{T}\)
- \(250\ \mu\text{T}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Write the expression for magnetic field on the axis of a circular loop.
The magnetic field at a point on the axis of a circular loop is \[ B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}, \] where \[ R=6\ \text{cm}, \] and \[ x=8\ \text{cm}. \] Given, \[ B=27\ \mu\text{T}. \]
Step 2: Write the magnetic field at the centre of the loop.
At the centre, \[ B_0=\frac{\mu_0 I}{2R}. \] Dividing the two expressions, \[ \frac{B}{B_0} = \frac{R^3}{(R^2+x^2)^{3/2}}. \] Therefore, \[ B_0 = B\frac{(R^2+x^2)^{3/2}}{R^3}. \]
Step 3: Substitute the given values.
\[ R^2+x^2 = 6^2+8^2 = 36+64 = 100. \] Hence, \[ (R^2+x^2)^{3/2} = 100^{3/2} = 10^3 = 1000. \] Also, \[ R^3=6^3=216. \] Therefore, \[ B_0 = 27\times\frac{1000}{216}\ \mu\text{T}. \] \[ B_0 = \frac{27000}{216}\ \mu\text{T}. \] \[ B_0 = 125\ \mu\text{T}. \]
Step 4: Final conclusion.
Hence, the magnetic field at the centre of the circular loop is \[ \boxed{125\ \mu\text{T}} \] Therefore, the correct option is \[ \boxed{(2)} \]
The magnetic field at a point on the axis of a circular loop is \[ B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}, \] where \[ R=6\ \text{cm}, \] and \[ x=8\ \text{cm}. \] Given, \[ B=27\ \mu\text{T}. \]
Step 2: Write the magnetic field at the centre of the loop.
At the centre, \[ B_0=\frac{\mu_0 I}{2R}. \] Dividing the two expressions, \[ \frac{B}{B_0} = \frac{R^3}{(R^2+x^2)^{3/2}}. \] Therefore, \[ B_0 = B\frac{(R^2+x^2)^{3/2}}{R^3}. \]
Step 3: Substitute the given values.
\[ R^2+x^2 = 6^2+8^2 = 36+64 = 100. \] Hence, \[ (R^2+x^2)^{3/2} = 100^{3/2} = 10^3 = 1000. \] Also, \[ R^3=6^3=216. \] Therefore, \[ B_0 = 27\times\frac{1000}{216}\ \mu\text{T}. \] \[ B_0 = \frac{27000}{216}\ \mu\text{T}. \] \[ B_0 = 125\ \mu\text{T}. \]
Step 4: Final conclusion.
Hence, the magnetic field at the centre of the circular loop is \[ \boxed{125\ \mu\text{T}} \] Therefore, the correct option is \[ \boxed{(2)} \]
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