SELECT * FROM question_mgmt as q WHERE id=2315 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=8 AND chapterId=119 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th physics Magnetism and Matter a bar magnet of magnetic moment 1 5 j t 1 nbsp lie

Question:

A bar magnet of magnetic moment 1.5 J T-1 lies aligned with the direction of a uniform magnetic field of 0.22 T.

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?

Answer:

 (a)Magnetic moment, M= 1.5 J T-1

Magnetic field strength, B= 0.22 T

(i)Initial angle between the axis and the magnetic field, θ1= 0°

Final angle between the axis and the magnetic field, θ2= 90°

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

W = - MB ( cos θ2 - cos θ1)
  = -1.5 x 0.22 ( cos 90° -  cos 0°)
  = -0.33 (0 - 1) 
  = 0.33 J

(ii)Initial angle between the axis and the magnetic field, θ1= 0°

Final angle between the axis and the magnetic field, θ2= 180°

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

W = - MB (cos θ2 - cos θ1)
  = -1.5 x 0.22 (cos 180° - cos 0°)
  = -0.33 (-1 - 1) 
  = 0.66 J

(b)For case (i):  θ = θ2 = 90°

∴Torque, T = MB sinθ
= 1.5 x 0.22 sin 90°
=0.33 J

Forcase (ii):  θ = θ2 = 180°

∴Torque, T = MB sin θ
=MB sin 180° = 0 


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SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='8' AND ex_no!=0 AND status=1 and id!=2315 ORDER BY last_viewed_on desc limit 0,10

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