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Question: A bar magnet when placed at an angle of \[30^\circ \] to the direction of magnetic field of inductio...

A bar magnet when placed at an angle of 30∘30^\circ to the direction of magnetic field of induction of 5× 10−5T5 \times {\text{ }}{10^{ - 5}}T, experiences a moment of a couple 2.5× 10−6N−m2.5 \times {\text{ }}{10^{ - 6}}N - m, If the length of the magnet is 5cm5cm its pole strength is
A.2× 10−2Am2 \times {\text{ }}{10^{ - 2}}Am
B.5× 10−2Am5 \times {\text{ }}{10^{ - 2}}Am
C.2Am2Am
D.5Am5Am

Explanation

Solution

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\tau {\text{ }} = {\text{ }}MB{\text{ }}sin{\text{ }}\theta
Pole strength of bar magnet, m=MLm = \dfrac{M}{L}

Complete answer:
Given that,
The angle of rotation, θ = 30∘\theta {\text{ }} = {\text{ }}30^\circ
magnetic field of induction, B = 5× 10−5TB{\text{ }} = {\text{ }}5 \times {\text{ }}{10^{ - 5}}T
Torque, τ = 2.5× 10−6N−m\tau {\text{ }} = {\text{ }}2.5 \times {\text{ }}{10^{ - 6}}N - m
Length of the magnet, L=5cm=0.05mL = 5cm = 0.05m
The net torque acting on a bar magnet is given as
τ = MB sin θ\tau {\text{ }} = {\text{ }}MB{\text{ }}sin{\text{ }}\theta
Substituting the given values of τ, B, and θ we get
2.5× 10−6= M × 5× 10−5× sin 302.5 \times {\text{ }}{10^{ - 6}} = {\text{ }}M{\text{ }} \times {\text{ }}5 \times {\text{ }}{10^{ - 5}} \times {\text{ }}sin{\text{ }}30
M=2.5×10−6sin30×5×10−5M = \dfrac{{2.5 \times {\text{}}{{10}^{ - 6}}}}{{sin30 \times 5 \times {\text{}}{{10}^{ - 5}}}}
M=2.5×10−60.5×5×10−5M = \dfrac{{2.5 \times {\text{}}{{10}^{ - 6}}}}{{0.5 \times 5 \times {\text{}}{{10}^{ - 5}}}}
∴M =0.1 A m2\therefore M{\text{ }} = 0.1{\text{ }}A{\text{ }}{m^2}
Now the pole strength of a magnet is given as
M=mLM = mL
0.1 =m× 0.050.1{\text{ }} = m \times {\text{ }}0.05
∴m=0.10.05=2×10−3=2mA\therefore m = \dfrac{{0.1}}{{0.05}} = 2 \times {10^{ - 3}} = 2mA

Option C is correct among all.

Note:
In physics, torque is considered as a rotational analog of force, and it is expressed as τ⃗=r⃗×F⃗\vec \tau = \vec r \times \vec 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.