Question

Two circular coils made from same wire but radius of 1st coil is twice that of 2nd coil. If magnetic field at their centres is same then ratio of potential difference applied across them is ($1^{st}$ to 2nd coil)

• A. 2
• B. 3
• C. 4
• D. 6

Hint:

Physics Tip : When you double the radius of a coil, you need twice the current to keep the magnetic field constant, and you also have twice the resistance because the wire is twice as long.

Answer 1

The Correct Option is C

Concept: Physics (Magnetism) - Magnetic Field of a Current Loop and Ohm's Law.

Step 1: Relate the currents through the magnetic field. The magnetic field $B$ at the center is $B = \frac{\mu_{0}I}{2r}$. Given $B_1 = B_2$ and $r_1 = 2r_2$: $$\frac{I_{1}}{2r_{1}} = \frac{I_{2}}{2r_{2}} \implies \frac{I_{1}}{I_{2}} = \frac{r_{1}}{r_{2}} = 2$$

Step 2: Relate the resistances. Resistance $R$ is proportional to the length of the wire ($l = 2\pi r$): $$\frac{R_{1}}{R_{2}} = \frac{r_{1}}{r_{2}} = 2$$

Step 3: Calculate the ratio of potential differences. Using Ohm's Law ($V = IR$): $$\frac{V_{1}}{V_{2}} = \frac{I_{1} \cdot R_{1}}{I_{2} \cdot R_{2}} = 2 \times 2 = 4$$ $$ \therefore \text{The ratio of potential difference is 4.} $$