This question asks for the specific distance on the axis of a circular coil where the magnetic field strength is reduced to one twenty-seventh (1/27th) of its value at the center.
Calculating Magnetic Field on Coil Axis
The magnetic field ($B_x$) on the axis of a circular coil with radius $R$, carrying current $I$, at a distance $x$ from the center is given by the formula:
$ B_x = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} $
The magnetic field ($B_0$) at the center of the coil (where $x=0$) is:
$ B_0 = \frac{\mu_0 I}{2R} $
Determining the Required Distance
We are given the condition that the magnetic field on the axis is $\frac{1}{27}$th of the field at the center:
$ B_x = \frac{1}{27} B_0 $
Substituting the formulas:
$ \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} = \frac{1}{27} \left( \frac{\mu_0 I}{2R} \right) $
Simplify by cancelling common terms ($\mu_0 I / 2$):
$ \frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{27R} $
Rearrange the equation:
$ 27 R^3 = (R^2 + x^2)^{3/2} $
To solve for $x$, raise both sides to the power of $\frac{2}{3}$:
$ (27 R^3)^{2/3} = \left( (R^2 + x^2)^{3/2} \right)^{2/3} $
$ (3^3 R^3)^{2/3} = R^2 + x^2 $
$ 3^2 R^2 = R^2 + x^2 $
$ 9 R^2 = R^2 + x^2 $
Now, isolate $x^2$:
$ x^2 = 9 R^2 - R^2 $
$ x^2 = 8 R^2 $
Finally, take the square root to find $x$:
$ x = \sqrt{8 R^2} $
$ x = \sqrt{8} R $
$ x = 2\sqrt{2} R $
Therefore, the distance from the center on the axis where the magnetic induction is $\frac{1}{27}$th of the value at the center is $2\sqrt{2} R$. This corresponds to Option A.
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Important Questions from Electricity and Magnetism
Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then
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B. The acceleration of the blocks is $a = \frac{kx}{M+m}$ (x = displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is $\frac{\mu m|x|}{M+m}$
D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu (M+m)g}{k}$
E. Maximum frictional force can be $\mu(M + m) g$.
Choose the correct answer from the options given below:
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