a bar magnet of magnetic moment 6j/t isaligned at 60° with a uniform external magnetic field of 0.44 T.calculate (a) the work done in turning the magnet to aligen its magnet moment (i) normal to the magnetic field,(ii)opposite to the magnetic field and (b) the torque on the magnet in a Final orientation in case of (ii)
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Answered by
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Formula of work done is given by
W = - MB[cosθf - cosθi]
Here M is the magnetic moment ,
B is the magnetic field
θf is final angle subtended by bar magnet with magnetic field
θi is the initial angle subtended by bar magnet with magnetic field.
(a) Normal to the magnetic field
θf = 90° , θi = 60° , M = 6J/T and B = 0.44T
so, work done = - 6 × 0.44 [ cos90° - cos60°]
= - 2.64 × [ 0 - 1/2 ] = 1.32 J
(b) opposite to the magnetic field
θf = 180° , θi = 60°
∴ work done = - 6 × 0.44 [ cos180° - cos60° ]
= -2.64 × (-1 - 1/2) = 2.64 × 3/2 = 1.32 × 3 = 3.96 J
W = - MB[cosθf - cosθi]
Here M is the magnetic moment ,
B is the magnetic field
θf is final angle subtended by bar magnet with magnetic field
θi is the initial angle subtended by bar magnet with magnetic field.
(a) Normal to the magnetic field
θf = 90° , θi = 60° , M = 6J/T and B = 0.44T
so, work done = - 6 × 0.44 [ cos90° - cos60°]
= - 2.64 × [ 0 - 1/2 ] = 1.32 J
(b) opposite to the magnetic field
θf = 180° , θi = 60°
∴ work done = - 6 × 0.44 [ cos180° - cos60° ]
= -2.64 × (-1 - 1/2) = 2.64 × 3/2 = 1.32 × 3 = 3.96 J
sumangill135:
let me explain -2.64 × (-1-1/2)
2.64×3/2
plz help me
-2.64 X (-1-1/2)
-2.64 X ((-2/2)-1/2)
-2.64 X ((-2-1)/2)
-2.64 X (-3/2)
Hope u understand.
thx
Ur most welcome dear
Answered by
1
(a) (i)
M = 6 J/T
Q1 = 60º Q2 = 90º
B = 0.44T
W = MB(cosQ1 - cosQ2)
= 6 x 0.44(cos60º - cos90º)
= 6 x 0.44 x ½
= 1.32 J
(a) (ii)
Q1 = 60º Q2 = 180º (opposite to magnetic field)
W = MB(cosQ1 - cosQ2)
= 6 x 0.44(cos60º - cos90º)
= 6 x 0.44 x (½ - (-1))
= 6 x 0.44 x 3/2
= 3.96 J
(b) = M X B
= MB sin Ø
= 6 x 0.44 x sin 180º
= 6 x 0.44 x 0
=0
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