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New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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

Explanation- dW=F.dl=0

As dl = vdt

dW= Fvdt

dW= f.v=0

New answer posted

10 months ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

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

When initially K1 is closed and K2 is open, the capacitors C1 and C2 connected in series with battery

Q=CV=C (C1C2/C1+C2) = 6 * 3/6+3=18 μ C

Now K1 was opened and K2 was closed, the battery and capacitor C, are disconnected from the circuit .

Here charge being shared by both capacitor so a common potential will develop.

V=C1V1+C2V2/C1+C2 = 18/3+3=3V

Q2=3CV=3 * 3=9μC

Q3=3CV=3 * 3=9μC

New answer posted

10 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

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

Explanation- (a) Suppose the five wires A, B, C, D and E be perpendicular to the plane of paper at locations as shown in figure.

Thus, magnetic field induction due to five wires will be represented by various sides of a

closed pentagon in one order, lying in the plane of paper. So, its value is zero.

(b) Since, the vector sum of magnetic field produced by each wire at O is equal to 0.

Therefore, magnetic induction produced by one current carrying wire is equal in

magnitude of resultant of four wires and opposite in direction.

Therefore, the field if current in one of the wi

...more

New answer posted

10 months ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

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

As we know that,   σ =charge/Area

Q1= σ (4 π r 2)

Q2= σ (4 π ( 2 r ) 2)= 4Q1

when they come in contact with each other potential being same and charge will be conserved.

So net charge is 5Q1= 5 ( σ (4 π r 2)

Also V1=V2

KQ1/R=KQ2/2R  so Q1=Q2/2

Q1= 5 3 ( σ (4 π r 2) and Q2= 10 3 ( σ (4 π r 2)

σ =5 σ /3  and σ  =5 σ /6

New answer posted

10 months ago

0 Follower 4 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

10 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

10 months ago

0 Follower 1 View

New answer posted

10 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

10 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- 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

10 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- 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

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