9. A ring of radius R having uniformly distributed charge () is
mounted on a rod suspended by two identical strings. The
tension in strings in equilibrium is To Now, a vertical
magnetic field is switched on and ring is rotated at constant
angular velocity co. Find the maximum co with which the ring
can be rotated if the strings can withstand a maximum tension
of 3T12
(2003.4M)
Tot
To
11
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Answer:
Answer
In equilibrium: 2T
o
=mg
or T
o
=
2
mg
...(i)
Magnetic moment, M=iA=(
2π
ω
Q)(πR
2
)
τ=MBsin90
o
=
2
ωBQR
2
Let T
1
and T
2
be the tensions in the two strings when magnetic field is switched on (T
1
>T
2
).
For translational equilibrium of ring is vertical direction,
T
1
+T
2
=mg ...(ii)
For rotational equilibrium,
(T
1
−T
2
)
2
D
=τ=
2
ωBQR
2
or T
1
−T
2
=
2
ωBQR
2
...(iii)
Solving equations (ii) and (iii) we have
T
1
=
2
mg
+
2D
ωBQR
2
As T
1
>T
2
and maximum values of T
1
can be
2
3T
o
, we have
2
3T
o
=T
o
+
2D
ω
max
BQR
2
∴ω
max
=
BQR
2
DT
o
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