Physics Ncert Solutions Class 12th

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New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- iG (G) = (I1-IG) (S1+S2+S3) for I1= 10mA

iG (G+S1) = (I2-IG) (S2+S3) for I2= 100mA

iG (G+S1+S2) = (I3-IG) (S3) for I3= 1A

S1= 1W, S2= 0,1W and S3= 0.01W

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1

(b) B.dl=μoI but when L  

So B  1/r3

(c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.

 

B= μ0IR22 (Z2+R2)3/2

Z=Rtan θ

dz=Rsec2 θdθ

-Bzdz=μ0I2-π/2π/2cosθdθ = μ0I

New question posted

4 months ago

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New answer posted

4 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Potential energy of system of charges be algebraic sum

U = 1 4 π ? ? ( q q r - q q r - q q r )

9 * 109/10-15 (1.6 * 10 -19)2 (1/3)2- (2/3) (1/3)- (2/3) (1/3)

-7.68 * 10-14J

-7.68 * 10-14 * 1.6 * 10 -19= 0.48Mev

New answer posted

4 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1

(b) B.dl=μoI but when L  

So B  1/r3

(c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.

 

B= μ0IR22 (Z2+R2)3/2

Z=Rtan θ

dz=Rsec2 θdθ

-Bzdz=μ0I2-π/2π/2cosθdθ = μ0I

New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as we know magnetic moment = nIA

For equilateral triangle M= nIA= 4I ( 34a2 )

M= Ia23

For square, n=3 so total length of wire is 12a

M= nIA= 3I (a2) = 3Ia2

For regular hexagon of side a, n=2 so total length = 12a

M= nIA=2 ( 634a2 )= 3 3 a2I

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- Since, B is along the x-axis, for a circular orbit the momenta of the two particles are in the y-z plane. Let P1 and P2 be the momentum of the electron and positron, respectively. Both traverse a circle of radius R of opposite sense. Let P1 make an angle? with the y-axis P2 must make the same angle.

The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at Ce and of the positron at Cp . The coordinates of Ce is

Ce= (0, -Rsin θ, Rcosθ )

Cp= (0, -Rsin θ, -32R-Rcosθ )

The circles of the two

...more

New answer posted

4 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- the thicker wire has a resistance R, then the other wire has a resistance @R as the wires are of the same material but different area

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – for equilibrium balance net torque should be zero

Mgl= Wcoill

500gl = Wcoill

Wcoil= 500 * 9.8N

Taking moment of force about mid point then magnetic field

Mgl+mgl=Wcoill+IBlsin90

Mgl=BILl

M= 0.2 * 4.9 * 1 * 10 - 2 9.8 = 10 -3kg= 1g

New question posted

4 months ago

0 Follower 7 Views

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