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Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What ratio of potential differences (in volts) should be applied across them so that the magnetic field at their centres is the same -
A
3
B
4
C
6
D
2
Detailed Solution
Let $r_1$ and $r_2$ be the radii of coils 1 and 2, and $B_1$ and $B_2$ the magnetic inductions at their centres.
$B_1 = \dfrac{\mu_0 I_1}{2r_1}$ and $B_2 = \dfrac{\mu_0 I_2}{2r_2}$
Since $B_1 = B_2$ and $r_1 = 2r_2$, therefore $I_1 = 2I_2$.
The coils are made from the same wire, so the resistance is proportional to the length $2\pi r$ of the wire: $R \propto r$.
If $R_1$ and $R_2$ are the resistances of coils 1 and 2, then $R_1 = 2R_2$.
If $V_1$ and $V_2$ are the potential differences across them: $\dfrac{V_1}{V_2} = \dfrac{I_1 R_1}{I_2 R_2} = \dfrac{(2I_2)(2R_2)}{I_2 R_2}$
$\dfrac{V_1}{V_2} = 4$
$B_1 = \dfrac{\mu_0 I_1}{2r_1}$ and $B_2 = \dfrac{\mu_0 I_2}{2r_2}$
Since $B_1 = B_2$ and $r_1 = 2r_2$, therefore $I_1 = 2I_2$.
The coils are made from the same wire, so the resistance is proportional to the length $2\pi r$ of the wire: $R \propto r$.
If $R_1$ and $R_2$ are the resistances of coils 1 and 2, then $R_1 = 2R_2$.
If $V_1$ and $V_2$ are the potential differences across them: $\dfrac{V_1}{V_2} = \dfrac{I_1 R_1}{I_2 R_2} = \dfrac{(2I_2)(2R_2)}{I_2 R_2}$
$\dfrac{V_1}{V_2} = 4$