Question

19
A short bar magnet placed with its axis inclined at 0 to 30° the external magnetic find of
800 G acting horizontally experiences a torque of 0.016 N-m. Calculate (1) magnetic moment of the
magnet (2) Work done in moving the dipole up 0 to 90°.​

Answer 1

Solution

$torque(T)=magnetic field(B)×magnetic \: moment[P(m)]×sin(theta)</p><p></p><p>$

=> T=P(m)Bsin∅

=>P(m)=T/(B sin∅)

now..given...

B=800G=0.08 T

∅=30°=>sin30°=>0.5

T=0.016N-m

therefore....

P(m)=0.016/(0.08*0.5)

=>P(m)=O.4 A.m²

now work done in moving the dipole from 0° to 90°.

W=P(m)B(1-cos∅)

=>W=(0.4)(0.08)(1-cos 90°)

=>W=0.4×0.08×1=0.032 wb-A

=>W=0.032 wb-A

Answer 2

Solution

torque(T)=magnetic field(B)×magnetic \: moment[P(m)]×sin(theta)torque(T)=magneticfield(B)×magneticmoment[P(m)]×sin(theta)

=> T=P(m)Bsin∅

=>P(m)=T/(B sin∅)

now..given...

B=800G=0.08 T

∅=30°=>sin30°=>0.5

T=0.016N-m

therefore....

P(m)=0.016/(0.08*0.5)

=>P(m)=O.4 A.m²

now work done in moving the dipole from 0° to 90°.

W=P(m)B(1-cos∅)

=>W=(0.4)(0.08)(1-cos 90°)

=>W=0.4×0.08×1=0.032 wb-A

=>W=0.032 wb-A