Physics Magnetism and Matter
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New answer posted
a month agoContributor-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 agoContributor-Level 10
So,
so only one option is correct i.e. 18.4°
Angle rotated should be
Answer: (A) 18.4°
New answer posted
a month agoContributor-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 agoContributor-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 agoContributor-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.
New answer posted
a month agoContributor-Level 10
In ferromagnetic material, below Curie's temperature, a domain is defined as macroscopic region with zero magentisation.
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