Solveeit Logo

Question

Question: A coil of radius \(R\) carries current \({i_1}\). Another concentric coil of radius \[r(r \ll R)\] c...

A coil of radius RR carries current i1{i_1}. Another concentric coil of radius r(rR)r(r \ll R) carries current i2{i_2}. Places of two coils are mutually perpendicular and both the coils are free to rotate about common diameter. If the maximum kinetic energy of smaller coil when both are released is μ0i1i2πr2MRX(MR2+mr2)\dfrac{{{\mu _0}{i_1}{i_2}\pi r_{}^2MR}}{{X\left( {MR_{}^2 + mr_{}^2} \right)}}. Find XX?
Masses of coils are MM and mm, respectively.

Explanation

Solution

To solve this question think about the conservation of the angular momentum. By finding the inertia, angular velocity, potential energy we are able to find the solution for this question.

Formula used:
I1ω1 = I2ω2{I_1}{\omega _1}{\text{ = }}{I_2}{\omega _2}
12I1ω12+12I2ω22=U\dfrac{1}{2}{I_1}\omega _1^2 + \dfrac{1}{2}{I_2}\omega _2^2 = U
U=MBcosθU = MB\cos \theta

Complete step by step answer:
We can find the solutions by understanding the laws of conservation of angular momentum and energy.
The law of conservation of energy states that the energy can be neither created nor can it be destroyed.
Given that a coil of radius RR carries current i1{i_1} and another concentric coil of radius r(rR)r(r \ll R) carries current i2{i_2}. The Mass of the coil with radius RR is MM and mass of the coil with radius rr is mm.
Also, places of two coils are mutually perpendicular and both the coils are free to rotate about common diameter as shown in the diagram.

The magnetic induction at centre due to current in larger coil, B=μ0i12RμB = \dfrac{{{\mu _0}{i_1}}}{{2R}}{\mu _ \circ }
Where
BB is the magnetic induction.
μ{\mu _ \circ }is the permeability of free space.
RR is the radius
Also, magnetic moment is given as M=I×AM = I \times A
Where,
MM is the magnetic moment
II is the current
Hence for smaller coil magnetic moment M=πr2i2M = \pi r_{}^2{i_2}
Initially when the coils are perpendicular to each other the angle between MM and BB is 90{90^ \circ } hence potential energy is given as
U1=MBcos90=0{U_1} = - MB\cos {90^ \circ } = 0
When the coils become coplanar the angle between MM and BB is 0{0^ \circ } hence potential energy U2 is given as
U2=MBcos0=MB{U_2} = - MB\cos {0^ \circ } = - MB
Therefore, decrease in the potential energy, U=U1U2U = {U_1} - {U_2}
U=0(MB)\Rightarrow U = 0 - ( - MB)
U=MB=μ0i1i2πr22R\Rightarrow U = MB = \dfrac{{{\mu _0}{i_1}{i_2}\pi r_{}^2}}{{2R}}
This decrease in potential energy is converted into kinetic energy.
Also, kinetic energy becomes maximum when the coils are coplanar.
Let ω1{\omega _1} and ω2{\omega _2}be angular velocity and I1{I_1} I2{I_2} be the moment of inertia of larger and smaller coil respectively when the coils become coplanar.
Also I1=MR2{I_1} = MR_{}^2and I2=mr2{I_2} = mr_{}^2
According to law of conservation of angular momentum I1ω1 = I2ω2{I_1}{\omega _1}{\text{ = }}{I_2}{\omega _2}
Hence we have ω1=I2ω2I1{\omega _1} = \dfrac{{{I_2}{\omega _2}}}{{{I_1}}}
Also, kinetic energy of rotating body is given as 12Iω2\dfrac{1}{2}{I_{}}\omega _{}^2
Hence, according to law of conservation of energy,
12I1ω12+12I2ω22=U\Rightarrow \dfrac{1}{2}{I_1}\omega _1^2 + \dfrac{1}{2}{I_2}\omega _2^2 = U
Using the value ω1=I2ω2I1{\omega _1} = \dfrac{{{I_2}{\omega _2}}}{{{I_1}}}
We have,
12I1(I2ω2I1)2+12I2ω22=U\Rightarrow \dfrac{1}{2}{I_1}\left( {\dfrac{{{I_2}{\omega _2}}}{{{I_1}}}} \right)_{}^2 + \dfrac{1}{2}{I_2}\omega _2^2 = U
U=12I1ω22(I1I2+1)\Rightarrow U = \dfrac{1}{2}{I_1}\omega _2^2\left( {\dfrac{{{I_1}}}{{{I_2}}} + 1} \right)
12I1ω22=UI1I1+I2\Rightarrow \dfrac{1}{2}{I_1}\omega _2^2 = \dfrac{{U{I_1}}}{{{I_1} + {I_2}}}
By substituting the values for UU, I1{I_1} and I2{I_2} we get,
12I1ω22=μoi1i2πr2MR2(MR2+mr2)\Rightarrow \dfrac{1}{2}{I_1}\omega _2^2 = \dfrac{{{\mu _o}{i_1}{i_2}\pi r_{}^2MR}}{{2(MR_{}^2 + mr_{}^2)}}
Where kinetic energy is 12I1ω22\dfrac{1}{2}{I_1}\omega _2^2
By comparing the equations μ0i1i2πr2MRX(MR2+mr2)\dfrac{{{\mu _0}{i_1}{i_2}\pi r_{}^2MR}}{{X\left( {MR_{}^2 + mr_{}^2} \right)}} and μoi1i2πr2MR2(MR2+mr2)\dfrac{{{\mu _o}{i_1}{i_2}\pi r_{}^2MR}}{{2(MR_{}^2 + mr_{}^2)}} we get X=2X = 2

Therefore X=2X = 2 is the required answer.

Note:
The law of conservation of angular momentum states that there will be no change in the angular momentum that occurs if no external torque is acting on it. The negative sign in the potential energy UU indicates that both the coils are moving in the direction opposite to each other.