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

Question

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 is the ratio of the potential difference applied across them so that the magnetic field at their centres is the same?

Accepted Answer

1. **Magnetic Field Condition:** The magnetic field $B$ at the center of a circular coil of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$ .
   For the fields to be equal ($B_1 = B_2$):
   $$ \frac{\mu_0 I_1}{2r_1} = \frac{\mu_0 I_2}{2r_2} \implies \frac{I_1}{r_1} = \frac{I_2}{r_2} \implies \frac{I_1}{I_2} = \frac{r_1}{r_2} $$
2. **Resistance Relationship:** The resistance $R$ of a wire is $R = ho \frac{L}{A}$. Since the coils are made of the same wire, resistivity $ho$ and cross-sectional area $A$ are constant. The length $L$ of the wire forming the coil is proportional to the radius ($L = 2\pi r 	imes N$). Assuming single-turn coils (or equal turns), resistance is proportional to radius: $R \propto r$.
   $$ \frac{R_1}{R_2} = \frac{r_1}{r_2} $$
3. **Potential Difference:** According to Ohm's Law, $V = IR$ .
   $$ \frac{V_1}{V_2} = \frac{I_1 R_1}{I_2 R_2} = \left( \frac{I_1}{I_2} ight) 	imes \left( \frac{R_1}{R_2} ight) $$
4. **Calculation:** Substituting the ratios derived above (given $r_1 = 2r_2$, so $r_1/r_2 = 2$):
   $$ \frac{V_1}{V_2} = (2) 	imes (2) = 4 $$
   Therefore, the ratio of potential differences is 4.

Step-by-Step Solution

Practice All PHYSICS Questions