Two concentric coils each of radius equal to 2pi cm are placed at right angles to each other. 3 A an

Q: Why are the magnetic fields perpendicular to each other?

The direction of the magnetic field at the center of a circular coil is always perpendicular to the plane of the coil. Since the two coils are placed in mutually perpendicular planes, their resulting magnetic field vectors will also be perpendicular to each other.

Q: What is the unit Wb/m²?

Weber per square meter ($\text{Wb/m}^2$) is the SI unit for magnetic flux density, which is exactly equivalent to the Tesla ($\text{T}$).

Q: Why did we convert the radius from cm to meters?

To ensure dimensional consistency. The permeability of free space $\mu_0$ is given in SI units ($\text{T}\cdot\text{m/A}$), so all lengths must be converted to meters to get the final answer in standard SI units.

Q: Can we just add the magnetic fields algebraically like $3 + 4 = 7$?

No, magnetic field is a vector quantity. Because the fields are pointing in different directions (specifically, at $90^\circ$ to each other), you must use vector addition to find the resultant magnitude.

Q: What if the currents were flowing in opposite directions?

Reversing the current in a coil reverses the direction of its magnetic field vector. However, since the two vectors are still perpendicular to each other, the magnitude of the resultant field $\sqrt{B_1^2 + B_2^2}$ remains exactly the same.

Question

Two concentric coils each of radius equal to $2\pi$ cm are placed at right angles to each other. 3 A and 4 A are the currents flowing in each coil respectively. The magnetic induction in $	ext{Wb/m}^2$ at the centre of the coils will be ($\mu_0 = 4\pi 	imes 10^{-7} 	ext{ Wb/Am}$)

Options:

(A) $12 	imes 10^{-5}$

(B) $10^{-5}$

(C) $5 	imes 10^{-5}$

(D) $7 	imes 10^{-5}$

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine two circular coils, both with the exact same radius, but placed perfectly perpendicular to each other. One lies flat, say in the $x-y$ plane, and the other stands upright in the $y-z$ plane. They share the same center. This is a classic problem of vector superposition in electromagnetism.

We need to find the net magnetic field at their common center. First, let's recall the magnetic field produced by a single circular coil at its center. The formula is given by:

$$B = \frac{\mu_0 I}{2R}$$

### Calculating Individual Fields

Let's calculate the magnetic field for the first coil, which carries a current of $3	ext{ A}$. We substitute $\mu_0$ as $4\pi 	imes 10^{-7}	ext{ T}\cdot	ext{m/A}$, the current $I_1$ as $3	ext{ A}$, and the radius $R$ as $0.02\pi	ext{ m}$.

$$B_1 = \frac{4\pi 	imes 10^{-7} 	imes 3}{2 	imes 0.02\pi}$$

Notice how the $\pi$ in the numerator and denominator beautifully cancel out. Calculating this gives us:

$$B_1 = 3 	imes 10^{-5} 	ext{ Wb/m}^2$$

By the right-hand rule, this field points along the $z$-axis. Now, let's look at the second coil, carrying $4	ext{ A}$. We use the exact same formula. The radius is identical, only the current changes to $4	ext{ A}$.

$$B_2 = \frac{4\pi 	imes 10^{-7} 	imes 4}{2 	imes 0.02\pi}$$

Following the same cancellation, we get:

$$B_2 = 4 	imes 10^{-5} 	ext{ Wb/m}^2$$

Because this coil is perpendicular to the first one, its magnetic field will also be perpendicular to the first field, pointing along the $x$-axis.

### Vector Addition for the Net Field

So, we have two magnetic field vectors at the center, $\vec{B}_1$ and $\vec{B}_2$, and they are exactly at $90^\circ$ to each other. To find the net magnetic field, we must add them vectorially using the Pythagorean theorem.

$$B_{	ext{net}} = \sqrt{B_1^2 + B_2^2}$$

Let's plug in our values. We need the square root of the sum of the squares of $3 	imes 10^{-5}$ and $4 	imes 10^{-5}$.

$$B_{	ext{net}} = \sqrt{(3 	imes 10^{-5})^2 + (4 	imes 10^{-5})^2}$$

You might recognize the classic $3-4-5$ right triangle here! Factoring out the $10^{-5}$, the square root of $3^2 + 4^2$ is simply $5$.

$$B_{	ext{net}} = 5 	imes 10^{-5} 	ext{ Wb/m}^2$$

**The net magnetic induction is $5 	imes 10^{-5} 	ext{ Wb/m}^2$.** Always remember, magnetic fields are vectors, and their geometry is just as important as their magnitude!

Select Answer:

Visualized Solution

$R = 2 π cm$

$B = \frac{μ _{0} I}{2 R}$

$B_{1} = \frac{4 π \times 1 0 ^{- 7} \times 3}{2 \times 0.02 π}$

$B_{1} = 3 \times 1 0^{- 5} Wb/m^{2}$

$B_{2} = \frac{4 π \times 1 0 ^{- 7} \times 4}{2 \times 0.02 π}$

$B_{2} = 4 \times 1 0^{- 5} Wb/m^{2}$

$B_{net} = B_{1}^{2} + B_{2}^{2}$

$B_{net} = (3 \times 1 0^{- 5})^{2} + (4 \times 1 0^{- 5})^{2}$

$B_{net} = 5 \times 1 0^{- 5} Wb/m^{2}$

$B_{net} = B_{1}^{2} + B_{2}^{2} + 2 B_{1} B_{2} cos θ$

The Sigma Insight: Biot-Savart Law

Visualizing the Setup

Calculating Individual Fields

Vector Addition for the Net Field

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=2π cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B=2Rμ0​I​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B1​=2×0.02π4π×10−7×3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B1​=3×10−5 Wb/m2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2​=2×0.02π4π×10−7×4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2​=4×10−5 Wb/m2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bnet​=B12​+B22​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bnet​=(3×10−5)2+(4×10−5)2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bnet​=5×10−5 Wb/m2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bnet​=B12​+B22​+2B1​B2​cosθ​