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A wheel of radius R having charge Q, uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. The rod is suspended by light inextensible strings and a magnetic field B is applied as shown in the figure. The initial tensions in the strings are T 0 . If the breaking tension of the strings are 3T 0 / 2 , find the maximum angular velocity ω 0 with which the wheel can be rotated

A wheel  of radius R having charge Q, uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. The rod is suspended by light inextensible strings and a magnetic field B is applied as shown in the figure. The initial tensions in the strings are T0. If the breaking tension of the strings are 3T0 / 2 , find the maximum angular velocity ω0 with which the wheel can be rotated

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
6 years ago
Hello Student,
Please find the answer to your question
When the ring is not rotating
Wt. of ring = Tension in string
mg = 2T0
∴ T0 = mg / 2 …(i)
When the ring is rotating, we can treat it as a current carrying loop. The magnetic moment of this loop
M = iA = Q / T πr2 = Q / 2π ω x πR2
This current carrying loop will create its own magnetic field which will interact with the given vertical magnetic field in such a way that the tensions in the strings will become unequal. Let the tensions in the strings be T1 and T2.
For translational equilibrium
T1 + T2 = mg …(ii)
Torque acting on the ring about the centre of ring
= x
Τ = M x B x sin 90°
= Q / 2π ω x πR2 x B = QωBR2 / 2
NOTE : For rotational equilibrium, the torque about the centre of ring should be zero.
∴ T1 x D / 2 – T2 x D / 2 = QωBR2 / 2
⇒ T1 – T2 = QωBR2 / D …(iii)
On solving (ii) and (iii), we get
T1 = mg / 2 + QωBR2 / 2D
But maximum tension is 3T0 / 2
∴ 3T0 / 2 = T­0 + Qωmax BR2 / 2D [∵ T0 = mg / 2]
∴ ωmax = DT0 / BQR2
Thanks
Deepak patra
askIITians Faculty

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