Question

.A bar magnet of moment of Inertia of 500gm cm^(2) makes 10 oscillations per minute in a horizontal plane.What is its magnetic moment of the horizontal component of earth's magnetic field is 0.36 gauss? Given: Moment of inertia I=500 gm cm^(2) frequency n=10 oscillation per minute =(10)/(60) oscillations per second.Time period T=6 sec By=0.36 gauss. Explain domain theory of ferromagnetism.​

Answer 1

Answer:

Answer

Time period is given by

T=2π

MB

I

T

2

=4π

2

×

MB

I

M=4π

2

×

T

2

B

I

Time period T=

20

15

=0.75 s

∴ M=4π

2

×

16π

2

×10

−5

×(0.75)

2

9×10

−5

M=

4×0.5625

9

M=

2.25

9

=4A−m

2

Answer 2

Given:

Inertia = 500 gcm²

Number of oscillations = 10

Magnetic field = 0.36

Moment of Inertia = 500 gm cm²

To Find:

Magnetic moment

Explanation:

n = 10 oscillations per minutes

n = 10/60 oscillations per sec

T = 1/n = 60/10 = 6 sec

B(H) = B = 0.36 gauss

       = 0.36 × 10⁻⁴

m = ?

T = 2π√(I/mB)

Squaring on both sides

T² = 4π²(I/mB)

m = [4×(3.14)²×500×10⁻⁷] / [(6)²×0.36×10⁻⁴

   =[4×(3.14)²×5×10⁴×10⁻⁷] / [36×36×10⁻⁴]

   = (3.142×3.142×5×10)/ 9×36

m = 1.24 Am²

Answer is 1.24 Am²