Physics Times 07.2019

(Kiana) #1

3.Sol:


1
6

10
2 10 g V

R R


   
5 10^4
  RV (R Rg V)

I R I I Sg g ( g)

6
3 6

2 10 10
20
10 2 10

S mA


 

 
 
 
 2 mA

A

SG
R
S G



3
20 10 10 20 10 3
A 10
R


     

3 3
3 3

10 1000 50 10
10 51 10

V
A
V

i
R
R
R

 
 
      
    
     

4

51
5 10




(^) RA (^0) 
' 51 10 (^31000)
i i  RV  
  
 
Measured resistance
4
' 1000 5 10 4 5 10 980.
m 51 51
R i
i


         
For calculating the motional emf across the
length of the wire, let us project wire such that
B V l, ,
 
becomes mutually orthogonal. Thus
0 0 1 0
y
d BV dy B V dy
L


   
     
   
0 0
0
1
L y
B V dy
L


   
    


    


0 0

1
1
1

B V L

 
   
  
emf in loop is proportional to L for given value
of .
for

(^) 0 ;  2 B V L0 0
0 0 0 0
1 4
2 ; 1
3 3
  BV L B V L
 
    
 
The length of the projection of the wire y = x
of length 2 L on the y-axis is L thus the
answer remain unchanged
4.Sol:
Just after closing of switch S 1 charge on
capacitors is zero.
Replace all capacitors with wire.
3.Sol:
4.Sol:
5 5
25mA
70 100 30 200
i  
 
Now S 1 is kept closed for long time and the
circuit is in steady state. Current does not flow
in the circuit.

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