JEE Mains Model Paper 12

Initially in equilibrium, T1+T2 = mg and T1 = T2 = T0
⇒ T0

mg

2

When the magnetic field is switched on, let the tensions in the left and right strings be T1′ and T2′ respectively.
Magnetic moment, M = iA =

Q

(2π∕ω)

× πR2
and torque τ = MB sin 90° =

Q

(2π∕ω)

× πR2B =

ωBQR2

2

For translational equilibrium T1′+T2′ = mg ... (1)
For rotational equilibrium (T1′−T2′) R =

ωBQR2

2

T1′−T2′ =

ωBQR

2

... (2)
From (1) and (2) we get T1′ =

mg

2

+

ωBQR

4

> T2′
Given, T1′ ≤

3T0

2

⇒

mg

2

+

ωmaxBQR

4

=

3T0

2

⇒ T0 +

ωmaxBQR

4

=

3T0

2

⇒ T0 =

ωmaxBQR

2

∴ ωmax =

2T0

RBQ