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New answer posted
10 months agoContributor-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 agoContributor-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 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 agoContributor-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
New answer posted
10 months agoContributor-Level 10
This is a long answer type question as classified in NCERT Exemplar
As we know that, =charge/Area
Q1= (4 2)
Q2= (4 2)= 4Q1
when they come in contact with each other potential being same and charge will be conserved.
So net charge is 5Q1= 5 ( (4 2)
Also V1=V2
KQ1/R=KQ2/2R so Q1=Q2/2
Q1= (4 2) and Q2= (4 2)
=5 /3 and =5 /6
New answer posted
10 months agoContributor-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 agoContributor-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) 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=
Z=Rtan
dz=Rsec2
=
New question posted
10 months agoNew answer posted
10 months agoContributor-Level 10
This is a long answer type question as classified in NCERT Exemplar
Potential energy of system of charges be algebraic sum
U =
9 109/10-15 (1.6 -19)2 (1/3)2- (2/3) (1/3)- (2/3) (1/3)
-7.68 10-14J
-7.68 10-14 -19= 0.48Mev
New answer posted
10 months agoContributor-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) 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=
Z=Rtan
dz=Rsec2
=
New answer posted
10 months agoContributor-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 ( )
M= Ia2
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 ( )= 3 a2I
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