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A bar magnet when placed at an angle of 30 to the direction of magnetic field of induction of 5× 105T, experiences a moment of a couple 2.5× 106Nm, If the length of the magnet is 5cm its pole strength is
A.2× 102Am
B.5× 102Am
C.2Am
D.5Am

Answer
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Hint: Torque is defined as a coupled force acting on a body that causes a body to rotate about its axis.
Example: While opening a lead we apply coupled force (parallel, equal but in opposite direction ) to rotate it about its center.
Pole strength is a scalar quantity and it is defined as the strength of a bar magnet and its ability to attract other magnetic material towards itself and magnetic moment of a bar magnet is directly proportional to pole strength of a magnet.
Formula Used:
The torque acting on a bar magnet, τ = MB sin θ
Pole strength of bar magnet, m=ML

Complete answer:
Given that,
The angle of rotation, θ = 30
magnetic field of induction, B = 5× 105T
Torque, τ = 2.5× 106Nm
Length of the magnet, L=5cm=0.05m
The net torque acting on a bar magnet is given as
τ = MB sin θ
Substituting the given values of τ, B, and θ we get
2.5× 106= M × 5× 105× sin 30
M=2.5×106sin30×5×105
M=2.5×1060.5×5×105
M =0.1 A m2
Now the pole strength of a magnet is given as
M=mL
0.1 =m× 0.05
m=0.10.05=2×103=2mA

Option C is correct among all.

Note:
In physics, torque is considered as a rotational analog of force, and it is expressed as τ=r×F.
Also, we must note that if we cut a bar magnet in half its pole strength will remain unaffected whereas its magnetic dipole moment will become half of its original intensity.
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