64:125
Magnetic Field Calculation for Circular Coil
This problem requires calculating the ratio of the magnetic field magnitude at the center of a circular coil ($B_1$) compared to the magnetic field magnitude at a specific distance along its axis ($B_2$). We are given the ratio between the axial distance '$x$' and the coil radius '$R$'.
Formulas for Magnetic Field
- Magnetic field at the center ($B_1$): For a circular coil of radius $R$ carrying current $I$, the magnetic field strength exactly at the center is given by:
$ B_1 = \frac{\mu_0 I}{2R} $
Here, $\mu_0$ represents the permeability of free space. - Magnetic field on the axis ($B_2$): The magnetic field strength at a distance $x$ from the center along the axis of the same coil is calculated using:
$ B_2 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} $
Applying the Given Ratio
The problem states the ratio of the axial distance to the radius:
$ \frac{x}{R} = \frac{3}{4} $
From this, we can express $x$ in terms of $R$: $x = \frac{3}{4}R$. This value is substituted into the formula for $B_2$.
Calculating the Ratio $B_2 / B_1$
- First, simplify the term $(R^2 + x^2)$ using $x = \frac{3}{4}R$:
$ R^2 + x^2 = R^2 + \left(\frac{3}{4}R\right)^2 = R^2 + \frac{9}{16}R^2 $
Combine the terms: $ R^2 + \frac{9}{16}R^2 = \frac{16R^2 + 9R^2}{16} = \frac{25R^2}{16} $
- Next, calculate the term $(R^2 + x^2)^{3/2}$:
$ \left(\frac{25R^2}{16}\right)^{3/2} = \left(\frac{5R}{4}\right)^3 = \frac{125R^3}{64} $
- Substitute this result back into the expression for $B_2$:
$ B_2 = \frac{\mu_0 I R^2}{2 \left(\frac{125R^3}{64}\right)} = \frac{\mu_0 I R^2 \cdot 64}{2 \cdot 125R^3} = \frac{32 \mu_0 I}{125R} $
- Finally, determine the ratio $\frac{B_2}{B_1}$:
$ \frac{B_2}{B_1} = \frac{\left(\frac{32 \mu_0 I}{125R}\right)}{\left(\frac{\mu_0 I}{2R}\right)} $
Simplify the division of fractions: $ \frac{B_2}{B_1} = \frac{32 \mu_0 I}{125R} \times \frac{2R}{\mu_0 I} = \frac{32 \times 2}{125} = \frac{64}{125} $
The ratio $\frac{B_2}{B_1}$ is $\frac{64}{125}$, which corresponds to the ratio 64:125.
Similar Questions
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
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
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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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
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
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:
A wire of length $25 \ m$ and cross-sectional area $5 \ mm^2$ having resistivity of $2 \times 10^{-6} \ \Omega \ m$ is bent into a complete circle. The resistance between diametrically opposite points will be
(DROPPED)
A parallel plate capacitor is filled equally(half) with two dielectrics of dielectric constants $\varepsilon_1$ and $\varepsilon_2$, as shown in figures. The distance between the plates is $d$ and area of each plate is $A$. If capacitance in first configuration and second configuration are $C_1$ and $C_2$ respectively, then $\frac{C_1}{C_2}$ is :
First Configuration
The electrostatic potential on the surface of uniformly charged spherical shell of radius $R = 10 \ cm$ is $120 \ V$. The potential at the centre of shell, at a distance $r = 5 \ cm$ from centre, and at a distance $r = 15 \ cm$ from the centre of the shell respectively, are:
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