AIIMS 2006 Physics Magnetic Field Due to Current MCQ Question

To solve the problem of finding the separation $H$ at which the magnetic fields created by a circular loop of wire and a long straight wire are equal and opposite at the center of the loop, we need to analyze the magnetic fields produced by both current-carrying conductors.

Step 1: Understanding the Magnetic Fields

1. Magnetic Field due to a Circular Loop: The magnetic field at the center of a circular loop of radius $R$ carrying current $I_\ell$ is given by the formula:

   $$B_{\text{loop}} = \frac{\mu_0 I_\ell}{2R}$$

   where $\mu_0$ is the permeability of free space, approximately $4\pi \times 10^{-7} \, \text{T m/A}$.

2. Magnetic Field due to a Long Straight Wire: The magnetic field at a distance $H$ from a long straight wire carrying current $I_s$ is given by:

   $$B_{\text{wire}} = \frac{\mu_0 I_s}{2\pi H}$$

Step 2: Setting the Magnetic Fields to be Equal

At the center of the loop, we want the total magnetic field to be zero. This means:

$$B_{\text{loop}} - B_{\text{wire}} = 0$$

or equivalently,

$$B_{\text{loop}} = B_{\text{wire}}$$

Substituting the expressions for the magnetic fields:

$$\frac{\mu_0 I_\ell}{2R} = \frac{\mu_0 I_s}{2\pi H}$$

Step 3: Simplifying the Equation

We can cancel $\mu_0$ from both sides (provided $\mu_0 eq 0$):

$$\frac{I_\ell}{2R} = \frac{I_s}{2\pi H}$$

Now, cross-multiplying gives us:

$$I_\ell \cdot 2\pi H = I_s \cdot 2R$$

Dividing both sides by $2\pi I_\ell$:

$$H = \frac{I_s R}{\pi I_\ell}$$

Step 4: Analyzing the Options

From our calculation, we find that the separation $H$ is:

$$H = \frac{I_s R}{\pi I_\ell}$$

Now, let's compare this result with the options provided:

• Option A: $\frac{I_s R}{I_\ell \pi}$ - This is exactly what we derived, so this option is correct.
• Option B: $\frac{I_s R}{I_\ell \pi}$ - This is the same as Option A but is written again, so it is also technically correct but redundant.
• Option C: $\frac{\pi I_s c}{I_\ell R}$ - This option introduces a speed of light $c$ which is irrelevant in this context, making it incorrect.
• Option D: $\frac{I_s \pi}{I_\ell R}$ - This option also has the wrong arrangement of variables and does not match our derived equation, so it is incorrect.

Conclusion

The correct answer is A: $H = \frac{I_s R}{I_\ell \pi}$. This value of $H$ ensures that the magnetic fields from both the circular loop and the long straight wire cancel each other out at the center of the circular loop.