Physics Magnetism and Matter

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Based on theory.

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

M = µ? NiA
Here
µ? = Relative permeability
N = Number of turns
i = Current
A = Area of cross section
M = µ? NiA = µ? nliA
M = µ? niV = 1000 (1000) (0.5) (10? ³) = 500 = 5 * 10²Am²

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

x = retentivity

y = coercivity

z = saturation magnetization

 

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

I010=I0cos? 2? \frac {I_0} {10} = I_0 \cos^2 \theta cos? ? =110=0.31<12which is 0.707\cos \theta = \frac {1} {\sqrt {10} = 0.31 < \frac {1} {\sqrt {2} \quad \text {which is 0.707}

So,

? >45? and90? ? ? <45? \theta > 45^\circ \quad \text {and} \quad 90^\circ - \theta < 45^\circ

so only one option is correct i.e. 18.4°

Angle rotated should be

=90? ? 71.6? =18.4? = 90^\circ - 71.6^\circ = 18.4^\circ

Answer: (A) 18.4°

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

M = 9.85*10? ² A/m²

I = 5*10? kg-m²

T = 5s/10osc = 0.5 sec

B =?

T = 2π√ (I/MB)

B = 4π²I/ (MT²) = (4*9.85*5*10? )/ (9.85*10? ² * 0.5²)

B = (20*10? )/ (10? ² * 0.25) = 800*10? = 0.8*10? ³ T = 8mT

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

We know, μ? = 1 + χ?
B? ∝ μ?
B ∝ μ
(ΔB/B? ) = (B-B? )/B? = (μ-μ? )/μ? = μ/μ? - 1 = μ? - 1 = χ?
% (ΔB/B) = χ? * 100 = 2.2 * 10? * 100 = 22/10?
(Note: The value for χ? in the source image appears to have a typo as 2.2 * 10? , it has been corrected to 2.2 * 10? to match the final answer.)

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

When a soft ferromagnetic substance is placed in external magnetic field, the size of domain lying in the opposite direction of external magnetic field increases while size of domain lying in the opposite direction of field decreases if field is weak. However, if field is strong then the domain rotate in the direction of external magnetic field due to strong torque.

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P
Payal Gupta

Contributor-Level 10

In ferromagnetic material, below Curie's temperature, a domain is defined as macroscopic region with zero magentisation.

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A
alok kumar singh

Contributor-Level 10

M = m l

ΔI=2l2sin30?

= l 2

M = m I / 2

= M / 2

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