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)
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)
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If two identical perpendicular vectors have magnitude $X$, their resultant is always $\sqrt{2}X$.
Updated On: Apr 30, 2026
- $\frac{\mu_0 N I}{2R}$
- $\frac{\mu_0 N I}{\sqrt{2}R}$
- $\frac{\mu_0 N I}{2\sqrt{2}R}$
- $\frac{\mu_0 N}{2}$
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The Correct Option is B
Solution and Explanation
Step 1: Magnetic Field of Single Coil
$B_1 = B_2 = \frac{\mu_0 N I}{2R}$
Step 2: Vector Direction
The coils are perpendicular, so their magnetic fields $\vec{B}_1$ and $\vec{B}_2$ at the common centre are also perpendicular.
Step 3: Resultant Field
$B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{2} B_1$
Step 4: Final Calculation
$B_{net} = \sqrt{2} \times \frac{\mu_0 N I}{2R} = \frac{\mu_0 N I}{\sqrt{2} R}$
Final Answer:(B)
$B_1 = B_2 = \frac{\mu_0 N I}{2R}$
Step 2: Vector Direction
The coils are perpendicular, so their magnetic fields $\vec{B}_1$ and $\vec{B}_2$ at the common centre are also perpendicular.
Step 3: Resultant Field
$B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{2} B_1$
Step 4: Final Calculation
$B_{net} = \sqrt{2} \times \frac{\mu_0 N I}{2R} = \frac{\mu_0 N I}{\sqrt{2} R}$
Final Answer:(B)
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