Physics, asked by paromita879, 10 months ago

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

Answers

Answered by NirmalPandya
0

Given :

  • Length of wire = L
  • Current carried by wire = i

To prove :

  • Magnitude of maximum torque = \frac{L^{2}iB }{4\pi }

Solution :

  • The wire is formed in a circular coil. Let r be the radius of the coil and the number of turns be N.  
  • L = 2\pi rN

     ∴ r = \frac{L }{2\pi N }

  • We know that, Torque on a current loop is given by :

        τ = NIAB sinα    ....(1)

  • Area (A) = πr²

                      = \pi (\frac{L}{2\pi N} )^{2}

                      =  \frac{L^{2} }{4\pi N^{2} }

  • Substituting the value of A in (1),

        τ = Ni(\frac{L^{2} }{4\pi N^{2} }) B sin\alpha

        τ = \frac{L^{2}IBsin\alpha  }{4\pi N}

  • 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 =  \frac{L^{2}iB }{4\pi }
  • 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 .

Answered by Fatimakincsem
0

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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