a bar magnet has poles of strength 48 Am, which are 25 cm apart.(i) What is the magnetic moment of the magnet? (ii) What torque is required to hold this magnet at an angle 30 degree with a uniform field of flux density 0.15 T?
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Given a bar magnet has poles of strength 48 Am, which are 25 cm apart.(i) What is the magnetic moment of the magnet? (ii) What torque is required to hold this magnet at an angle 30 degree with a uniform field of flux density 0.15 T?
- Given A bar magnet has poles of strength 48 Am, which are 25 cm apart. we need to find the force to rotate at angle 30 degree in this field.
- So we have Torque = MB sin theta ---------1
- So now M = m x L
- = 48 x 25 x 10^-2
- Substituting in eqn 1 we get
- So torque = 48 x 25 x 10^-2 x 0.15 x sin 30
- = 48 x 25 x 10^-2 x 0.15 x ½
- = 90 / 100
- = 0.9 N-m
- This is the required torque to rotate the angle 30 degree.
- Reference link will be
- https://brainly.in/question/14539505
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