Physics Times – July 2019

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.