Question

Two current carrying identical coils are kept as shown in figure. The magnetic field at centre 'O' is (N and R represent the number of turns and radius of each coil respectively, $\mu_0$ = permeability of free space)

• A. $\frac{\mu_0 N I}{2R}$
• B. $\frac{\mu_0 N I}{\sqrt{2}R}$
• C. $\frac{\mu_0 N I}{2\sqrt{2}R}$
• D. $\frac{\mu_0 N}{2}$

Hint:

If two identical perpendicular vectors have magnitude $X$, their resultant is always $\sqrt{2}X$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The figure shows two identical circular coils perpendicular to each other, with a common center. Each creates a magnetic field at the center directed along its axis.
Step 2: Key Formula or Approach:
1. Magnetic field at the center of a circular coil: $B = \frac{\mu_0 N I}{2R}$
2. Since the coils are perpendicular, their fields $\vec{B_1}$ and $\vec{B_2}$ are perpendicular.
3. Net magnetic field $B_{net} = \sqrt{B_1^2 + B_2^2}$
Step 3: Detailed Explanation:
Both coils are identical and carry the same current.
$B_1 = \frac{\mu_0 N I}{2R}$
$B_2 = \frac{\mu_0 N I}{2R}$
The vectors are at $90^\circ$ to each other.
\[ B_{net} = \sqrt{\left(\frac{\mu_0 N I}{2R}\right)^2 + \left(\frac{\mu_0 N I}{2R}\right)^2} \]
\[ B_{net} = \frac{\mu_0 N I}{2R} \sqrt{1^2 + 1^2} \]
\[ B_{net} = \frac{\mu_0 N I}{2R} \sqrt{2} \]
\[ B_{net} = \frac{\mu_0 N I}{\sqrt{2}R} \]
Step 4: Final Answer:
The resultant magnetic field at the center is $\frac{\mu_0 N I}{\sqrt{2}R}$.