A short bar magnet is placed in a uniform magnetic field of 0.25 T, with its axis at an angle of 30° with the field. If it experiences a torque of magnitude 4.5×10−2 N m, then its magnetic moment is0.36 N m T−10.25 N m T−10.5 N m T−11.25 N m T−1
Question
A short bar magnet is placed in a uniform magnetic field of 0.25 T, with its axis at an angle of 30° with the field. If it experiences a torque of magnitude 4.5×10−2 N m, then its magnetic moment is0.36 N m T−10.25 N m T−10.5 N m T−11.25 N m T−1
Solution
The torque (τ) experienced by a magnet in a magnetic field is given by the equation:
τ = μBsinθ
where: μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between the magnetic moment and the magnetic field.
We can rearrange this equation to solve for the magnetic moment (μ):
μ = τ / (Bsinθ)
Given that the torque (τ) is 4.5×10−2 N m, the magnetic field strength (B) is 0.25 T, and the angle (θ) is 30°, we can substitute these values into the equation:
μ = 4.5×10−2 N m / (0.25 T * sin(30°))
Calculating the sine of 30° gives us 0.5, so the equation becomes:
μ = 4.5×10−2 N m / (0.25 T * 0.5)
Solving this equation gives us:
μ = 0.36 N m T−1
So, the magnetic moment of the bar magnet is 0.36 N m T−1.
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