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Bar magnet of magnetic moment 200 A-m2 is suspended in a magnetic field of intensity 0.25N/Am. The couple required to deflect it through 30° is
- a)15 N-m
- b)20N-m
- c)25 N-m
- d)50 N-m
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Bar magnet of magnetic moment 200 A-m2is suspended in a magnetic field...
The torque experienced by a bar magnet in a magnetic field is given by the equation:
τ = mBsinθ
Where:
τ = torque (Nm)
m = magnetic moment of the bar magnet (A-m²)
B = magnetic field intensity (T)
θ = angle between the magnetic moment and the magnetic field (radians)
In this case, the magnetic moment of the bar magnet is given as 200 A-m² and the magnetic field intensity is given as 0.25 N/Am.
To find the torque required to deflect the bar magnet through 30°, we need to calculate the angle θ in radians:
θ = 30° × (π/180°) = 0.5236 radians
Now we can calculate the torque:
τ = (200 A-m²) × (0.25 N/Am) × sin(0.5236 radians)
= 25 Nm
Therefore, the couple required to deflect the bar magnet through 30° is 25 Nm.
τ = mBsinθ
Where:
τ = torque (Nm)
m = magnetic moment of the bar magnet (A-m²)
B = magnetic field intensity (T)
θ = angle between the magnetic moment and the magnetic field (radians)
In this case, the magnetic moment of the bar magnet is given as 200 A-m² and the magnetic field intensity is given as 0.25 N/Am.
To find the torque required to deflect the bar magnet through 30°, we need to calculate the angle θ in radians:
θ = 30° × (π/180°) = 0.5236 radians
Now we can calculate the torque:
τ = (200 A-m²) × (0.25 N/Am) × sin(0.5236 radians)
= 25 Nm
Therefore, the couple required to deflect the bar magnet through 30° is 25 Nm.
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Bar magnet of magnetic moment 200 A-m2is suspended in a magnetic field of intensity 0.25N/Am. The couple required to deflect it through 30° isa)15 N-mb)20N-mc)25 N-md)50 N-mCorrect answer is option 'B'. Can you explain this answer?
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