A small bar magnet experiences a 0.020 N m torque when the axis of the magnet is at 45 ∘ to a 0.10 T magnetic field. What is the magnitude of its magnetic dipole moment?

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A small bar magnet experiences a 0.020 N m torque when the axis of the magnet is at       45          ∘       to a 0.10 T magnetic field. What is the magnitude of its magnetic dipole moment?

garnboernl · Accepted Answer

The torque   τ represents the magnitude of the force multiplied by the distance between the axis and the line of the action. A current loop inside a magnetic fields experience a torque that is exerted by the magnetic force and the loop starts to rotate. For a square loop with side l.  τ  =  μ  B  sin  ⁡  θ...(1)Where   μ&amp;lt;  θ/math&amp;gt; represents the magnetic dipole moment of the loop, I is the current in the loop and  θ is the angle between the magnetic field and the magnetic dipole moment of the loop.When the dipole moment is parallel to the magnetic field, the torque is zero and it is maximum when the dipole moment is perpendicular to the magnetic field. Our target is to find   μ , so we rearrange equation (1) for   μ to be in the form  μ  =      τ          B      sin      ⁡      θ      Now, we plug the values for   τ , B and   θ into equation (2) to get   μ  μ  =      τ          B      sin      ⁡      θ                  0.020      N      ∗      m              (      0.10      T      )      sin      ⁡              45                  ∘                      =  0.28  A  ∗      m          2      Result:  μ  =  0.28  A  ∗      m          2

Haiphongum · Accepted Answer

We have to find magnitude of its magnetic dipole moment.Given values:  τ  =  0.02  N  mB=0.10T  θ  =      45          ∘      Net torque in uniform magnetic field.      τ    ¯    =      M    ¯    ×  B (M is magnetic dipole moment)  τ  =  M  B  sin  ⁡      45          ∘        M  =      2          B      sin      ⁡              45                  ∘                      M  =      0.02          0.1      ×              1                  2                      M  =  0.282  A      m          2      Result:  M  =  0.282  A      m          2

A small bar magnet experiences a 0.020 N m torque when the axis of the magnet is at 45^(circ) to a 0.10 T magnetic field. What is the magnitude of its magnetic dipole moment?

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