A length L of wire carries a current i. Show that if the wire is formed into a circular coil, then the maximum torque in a given magnetic field is developed when the coil has one turn only, and that maximum torque has the magnitude tau=L^2iB/4pi
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Given :
- Length of wire = L
- Current carried by wire = i
To prove :
- Magnitude of maximum torque =
Solution :
- The wire is formed in a circular coil. Let r be the radius of the coil and the number of turns be N.
∴
- We know that, Torque on a current loop is given by :
τ = NIAB sinα ....(1)
- Area (A) = πr²
=
=
- Substituting the value of A in (1),
τ =
τ =
- From this equation, we can see that τ is maximum when N will be minimum and sinα will be maximum.Minimum value of N will be 1 and maximum value of sinα=1 .Hence, we take N = 1 and sinα = 1
- ∴ τ =
- This proves that maximum torque will be developed when the coil has one turn only. Also, the coil will be in the plane of the magnetic field .
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τmax = L^2.I B / 4π
Hence proved that torque is maximum when number of turn are minimum.
Explanation:
Torque on the current loop is τ = NIABsinα --------(1)
If there are N turns of circular coil, each of radius r, then
L=2πrN
or r = L/2πN
Area of the coil, A = πr^2 =πL^2 / (4π^2 N^2) = L^2/(4πN^2)
Putting this value in (1) We get
τ = NI[L^2 / 4πN^2] Bsinα = L^2 I Bsinα / 4πN -------(2)
From equation (2) it is clear that τ is maximum if N is minimum
- That is the number of turns N = 1
- Torque is maximum if sinα = 1 and N = 1
∴ τmax = L^2.I B / 4π
Hence proved that torque is maximum when number of turn are minimum.
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