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
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
Show Hint
Magnetic flux is the number of magnetic field lines passing through a given closed surface.
6.6 x 10-9 Weber
9.1 x 10-11 Weber
6 x 10-11 Weber
3.3 x 10-11 Weber
The Correct Option is B
Approach Solution - 1
Let the current in the bigger loop be i2 and the smaller loop be i1
- ϕ1 is the flux due to the smaller loop at the bigger loop
- ϕ2 be flux due to a bigger loop at the smaller loop
The field due to the current loop 1 at an axial point -
\(B_{1}=\frac{\mu_{0}I_{1}R^{2}}{2\left(d^{2}+R^{2}\right)^{3 /2}}\)
Flux linked with the smaller loop 2 due to B1 is
\(\phi_{2}=B_{1}A_{2}=\frac{\mu_{0}I_{1}R^{2}}{2\left(d^{2}+R^{2}\right)^{3/ 2}}\pi r^{2}\)
The coefficient of mutual inductance between the loops is
\(M=\frac{\phi_{2}}{I_{1}}=\frac{\mu_{0}R^{2}\pi r^{2}}{2\left(d^{2}+R^{2}\right)^{3 / 2}}\)
Flux linked with bigger loop 1 is
\(\phi_{1}=MI_{2}=\frac{\mu_{0}R^{2}\pi r^{2}l^{2}}{2\left(d^{2}+R^{2}\right)^{3 / 2}}\)
Substituting the given values, we get
\(\phi_{1}=\frac{4\pi\times10^{-7}\times\left(20\times10^{-2}\right)^{2}\times\pi\times\left(0.3\times10^{-2}\right)^2\times2}{2\left[\left(15\times10^{-2}\right)^{2}+\left(20\times10^{-2}\right)^{2}\right]^{3/ 2}}\)
Option B is the correct answer, ϕ1 = 9.1 x 10-11 Weber
Approach Solution -2
Magnetic flux is the number of magnetic field lines passing through a given closed surface. Thus magnetic flux is the product of the magnetic field and its area.
Formula used:
Flux formula, ϕ = BA
- B = Magnetic field
- A = area.
\(B=\frac{μ_{0}IR^{2}}{2(R+x)^{\frac{3}{2}}}\) and \(A=πr^{2}\)
Complete Answer:
Circular loop radius, r = 0.3 cm = 3 x 10-3 m
Bigger circular loop radius, R = 20 cm = 0.2 m
Distance, x = 15 cm = 0.15 m
Current flowing, I = 2.0 A
To calculate flux, we first have to calculate B,
\(B=\frac{μ_{0}IR^{2}}{2(R+x)^{\frac{3}{2}}}=\frac{4π×10^{−7}×2×0.2^{2}}{2(0.2^{2}+0.15^{2})^{\frac{3}{2}}}=3.22×10^{−6}T\)
Then,
ϕ = BA = 3.22 x 10-6 x 3.14 x (3 x 10-3)2 = 9.1 x 10-11 Weber
Thus option (B) is correct.
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 x 10-11 Weber.
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Concepts Used:
Electromagnetic Induction
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
- When we place the conductor in a changing magnetic field.
- When the conductor constantly moves in a stationary field.
Formula:
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where
- e = induced voltage
- N = number of turns in the coil
- Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
- t = time
Applications of Electromagnetic Induction
- Electromagnetic induction in AC generator
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- Magnetic Flow Meter