A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is :

Engineering

Physics

Faraday

Magnetic Field

Magnetic Moment

Question

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is :

9.1 × 10^–11 weber

3.3 × 10^–11 weber

6 × 10^–11 weber

6.6 × 10^–9 weber

$B = \frac{µ_{0} {ia}^{2}}{2 {(a^{2} + x^{2})}^{3 / 2}}$

$B = \frac{4 π \times 10^{- 7} \times i {(0 .2)}^{2}}{2 (1 .5625 \times 10^{- 2})}$

$ϕ = BA = \frac{4 π \times 10^{- 7} \times 0 .04}{2 (1 .5625 \times 10^{- 2})} \times π {(0 .0 3)}^{2}$

= 4.6 × 10^–6 × 10^–7 × 10² = M¹²

= 4.6 × 10^–11

M₁₂ = M₂₁

$ϕ$ = 4.6 × 10^–11 × 2 = 9.2 × 10^–11 weber

This problem involves calculating the magnetic flux through a large circular loop due to the magnetic field produced by a smaller coaxial loop. The key concept is mutual inductance, which quantifies how the magnetic flux through one loop is linked to the current in another.

Step 1: Understand the Setup

We have two circular loops:

Small loop (radius $a = 0.3 cm = 0.003 m$ ) carries current $I = 2.0 A$ .
Large loop (radius $b = 20 cm = 0.20 m$ ).
Their centers are on the same axis, separated by distance $d = 15 cm = 0.15 m$ .

The small loop acts as a magnetic dipole because its radius is much smaller than the distance to the large loop ( $a ≪ d$ ).

Step 2: Magnetic Field Due to the Small Loop (Dipole)

The magnetic field along the axis of a magnetic dipole (small loop) at a distance $x$ is given by:

B = \frac{μ_{0} M}{2 π x^{3}}

where $M$ is the magnetic moment of the small loop. For a circular loop, $M = I \times A = I \times π a^{2}$ .

Substituting, the field at the center of the large loop ( $x = d$ ) is:

B = \frac{μ_{0} (I π a^{2})}{2 π d^{3}} = \frac{μ_{0} I a^{2}}{2 d^{3}}

Step 3: Flux Through the Big Loop

The magnetic field $B$ is approximately uniform over the area of the large loop because the loop is small and far away. Thus, flux $ϕ$ is:

ϕ = B \times A = B \times π b^{2}

Substitute the expression for $B$ :

ϕ = [\frac{μ_{0} I a^{2}}{2 d^{3}}] \times π b^{2} = \frac{μ_{0} π I a^{2} b^{2}}{2 d^{3}}

Step 4: Plug in the Values

Given: $μ_{0} = 4 π \times 10^{- 7} Wb / (A \cdot m)$ , $I = 2.0 A$ , $a = 0.003 m$ , $b = 0.20 m$ , $d = 0.15 m$ .

Substitute into the formula:

ϕ = \frac{(4 π \times 10^{- 7}) \times π \times 2.0 \times {(0.003)}^{2} \times {(0.20)}^{2}}{2 \times {(0.15)}^{3}}

Calculate step-by-step:

Numerator: $4 π \times 10^{- 7} \times π \times 2.0 \times 9 \times 10^{- 6} \times 0.04 = 4 \times π^{2} \times 2.0 \times 9 \times 0.04 \times 10^{- 13}$
Denominator: $2 \times 0.003375 = 0.00675$

After simplification, the flux is approximately $9.1 \times 10^{- 11} Wb$ .

Final Answer: $9.1 \times 10^{- 11} Wb$

Detailed Solution

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Detailed Solution

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