The magnetic field at the centre of a circular coil of radius r and carrying I is B. What is the magnetic field at a distance \(x = \sqrt{3}r\) from the centre, on the axis of the coil?
Understanding Magnetic Field of a Circular Coil
The question asks us to find the magnetic field on the axis of a circular coil at a specific distance, given the magnetic field at the centre of the coil. We need to use the standard formulas for the magnetic field produced by a current-carrying circular coil at its centre and on its axis.
Magnetic Field at the Centre of a Circular Coil
The magnetic field at the centre of a circular coil with radius \(r\) carrying a current \(I\) is given by the formula:
\[ B_{centre} = \frac{\mu_0 I}{2r} \]
We are given that this magnetic field at the centre is \(B\). So, \(B = \frac{\mu_0 I}{2r}\).
Magnetic Field on the Axis of a Circular Coil
The magnetic field on the axis of a circular coil with radius \(r\) carrying a current \(I\) at a distance \(x\) from the centre is given by the formula:
\[ B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}} \]
Calculating Magnetic Field at \(x = \sqrt{3}r\)
We need to find the magnetic field at a distance \(x = \sqrt{3}r\) from the centre, on the axis. We substitute \(x = \sqrt{3}r\) into the formula for \(B_{axis}\):
\[ B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + (\sqrt{3}r)^2)^{3/2}} \]
Simplify the term in the parenthesis:
\[ r^2 + (\sqrt{3}r)^2 = r^2 + 3r^2 = 4r^2 \]
Substitute this back into the \(B_{axis}\) formula:
\[ B_{axis} = \frac{\mu_0 I r^2}{2(4r^2)^{3/2}} \]
Now, evaluate \((4r^2)^{3/2}\):
\[ (4r^2)^{3/2} = (4^{1/2} \cdot (r^2)^{1/2})^3 = (2 \cdot r)^3 = 8r^3 \]
Substitute this back into the \(B_{axis}\) formula:
\[ B_{axis} = \frac{\mu_0 I r^2}{2(8r^3)} = \frac{\mu_0 I r^2}{16r^3} \]
Cancel out \(r^2\) from the numerator and denominator:
\[ B_{axis} = \frac{\mu_0 I}{16r} \]
Comparing \(B_{axis}\) with \(B_{centre}\)
We found \(B_{axis} = \frac{\mu_0 I}{16r}\) at \(x = \sqrt{3}r\) and we know \(B_{centre} = B = \frac{\mu_0 I}{2r}\). We can rewrite \(B_{axis}\) in terms of \(B\):
\[ B_{axis} = \frac{1}{8} \left( \frac{\mu_0 I}{2r} \right) \]
Since \(B = \frac{\mu_0 I}{2r}\), we have:
\[ B_{axis} = \frac{1}{8} B = \frac{B}{8} \]
Conclusion
The magnetic field at a distance \(x = \sqrt{3}r\) from the centre, on the axis of the coil, is \(B/8\), where \(B\) is the magnetic field at the centre.
Let's summarize the values:
| Location | Distance from Centre | Magnetic Field Formula | Magnetic Field Value |
|---|---|---|---|
| Centre | \(x=0\) | \(\frac{\mu_0 I}{2r}\) | \(B\) |
| On axis | \(x=\sqrt{3}r\) | \(\frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}\) | \(\frac{B}{8}\) |
Therefore, the magnetic field at \(x = \sqrt{3}r\) on the axis is \(B/8\).
Important Questions from Magnetic Field
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