CBSE Physics/ Lakshmikanta Satapathy/ Electromagnetism QA/ Magnetic field due to circular coil at center & on the axis/ Magnetic field due to Straight conductor/ Magnetic Lorentz force

1. Physics Helpline
L K Satapathy
Electromagnetism -1
Magnetic Field / Force
:
2
onI
Circularcoil atCenter B
R


 
2
3 2
2 2
:
2
onIR
Circularcoil onaxis B
R x



 :Magneticforce F q v B 
ur r ur

2. Physics Helpline
L K Satapathy
Electromagnetism -1
Q1. A circular coil of radius R carries a current I. The magnetic field at the center
of the coil is B. It is observed that, the magnetic field at a point P on the axis of the
coil is B/8. Determine the distance of point P from the center of the coil.
Answer :
At the center O of the coil of n-turns . . . (1)
2
onI
B
R


Let the distance OP = x .
 
2
3 2
2 2
. . . (2)
2
onIR
B
R x

 

 Magnetic field at P is given by
 
 
2
3 2
2 2 3
3 2
2 2
1
8
8 8 22
o onIR nIB
B R x R
RR x
 
       

 
2 3
2 2 3 2 2 2
8 4 3 3 [ ]R x R R x R x R Ans       
O P
x

3. Physics Helpline
L K Satapathy
Electromagnetism -1
Q2. An infinite straight conductor is carrying a current I .
A portion of it is bent into a circle of radius R as shown in
the figure. Determine the magnetic field at the centre of
the circle.
Answer :
Magnetic field at center O due to circular coil 1 . . . (1)
2
oI
B
R


Magnetic field at center O due to straight wire 2 . . . (2)
2
oI
B
R



Both are in the same direction  The two fields add up
 1 2 1 [ ]
2 2 2
o o oI I I
B B B Ans
R R R
  

 
      
O

4. Physics Helpline
L K Satapathy
Electromagnetism -1
Q3. A proton is accelerated until it is moving with a speed of 10 km/s. It is then
allowed to enter a region of magnetic field of induction 5T. Determine the largest
deflecting force that can be experienced by it .
Answer : Force on a charge q moving with speed v
inside a magnetic field B is given by
19 4 15
max 1.6 10 10 5 8 10 [ ]F N Ans 
      
 F q v B 
ur r ur
sinF qvB 
ur
 Magnitude of force is given by
maxF qvB
ur
 Maximum force is given by
Given : v = 10 km/s & B = 5T
19
1.6 10q C
 Also for a proton

5. Physics Helpline
L K Satapathy
Electromagnetism -1
Q4. A magnetic field of Tesla exerts a force of on a
particle having charge moving in the x-y plane. Determine the velocity of
the particle .
Answer :
3 ˆ(4 10 )k
 10ˆ ˆ(4 3 ) 10i j N
 
  . . . (1)F q v B 
ur r ur
Let the velocity of the particle be
10 9 3 ˆˆ ˆ: (4 3 ) 10 , 10 , (4 10 )Given F i j N q C B k T  
     
ur ur
ˆ ˆ( )x yv v i v j m s 
r
10 9 3 ˆˆ ˆ ˆ ˆ(1) (4 3 ) 10 10 ( ) (4 10 )x yi j v i v j k         
 
ˆˆ ˆ ˆ ˆ ˆ ˆ(400 300 ) ( ) (4 ) 4 ( ) 4 ( )x y x yi j v i v j k v j v i       
300 400
75 & 100
4 4
x yv v      
ˆ ˆ ˆ ˆ( ) ( 75 100 ) [ ]x yv v i v j i j m s Ans    
r
9
10 C

6. Physics Helpline
L K Satapathy
Electromagnetism -1
Q5. A stream of charged particles moving undeflected with uniform speed of
500m/s in a direction perpendicular to crossed electric and magnetic fields. The
magnitude of the electric field is 1000V/m. Determine the magnitude of the
magnetic field .
Answer :
& ( )e mF qE F q v B  
uur ur uur r ur
&e mF qE F qvB 
e mF F qE qvB  
1000
2 [ ]
500
E
B T Ans
v
   
: 500 & 1000Given v m s E V m 
[ Since v is  to both E & B ]

7. Physics Helpline
L K Satapathy
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