FORM 6 PHYSICS: ELECTROMAGNETISM PART 2

Solution

( i)   The  magnitude  of the  magnetic  field  inside  the  toroid is  given  by

2

B = 0.003T

(ii)The magnetic field outside the toroid is Zero. It is all inside the toroid.

    11.   A solenoid 1.5m long and 4cm in diameter possess 10 turnscm. A current of 5A is flowing through it. Calculate the  magnetic  induction

(i)   Inside  and

(ii)  At  one end  on the  axis  of the  solenoid

Solution
n = =  =

(i)    Inside the solenoid , the magnetic induction is given by

B =

                B = 4

B =

(ii)    At the  end  of the  solenoid  the  magnetic  induction is  given  by

     12.    (a)  How will the magnetic field intensity at the centre of a circular loop carrying current change, if the current through the coil is doubled and the radius of the coil is halved?

 (b) A long wire first bent in to a circular coil of one turn and then into a circular

          coil    of  smaller radius having  n  turns, if the  same  current passes in both the  cases, find  the  ratio of  magnetic  fields produced at the  centers in the  two cases.

(c) A and B are  two concentric coils of centre O and carry currents I_A and I_B  as shown in figure

                               

If the ratio of their radii is 1:2 and ratio of flux densities at O due to A and B is 1:3, find the value of

         Solution

(a)      Magnetic field at the centre of circular coil

         ( b)Suppose r is the radius of one turn coil and the r¹ is the radius of n-turn coil. Then

N

First case                                  Second case                                        

                                  

Solution
C.  Magnetic field at the centre of circular coil

     13.   A helium nucleus makes a full rotation in a circle of radius 0.8m in two seconds. Find the value of magnetic field at the centre of the circle.

Solution

The charge on helium nucleus

Q= e

Q=   1.6 X10^-19c

Current produced I =

I = 2 x 1.6 x 10^-19

2

I =1.6 x10^-19A

Magnetic field at the centre of the circle orbit of the helium is,

  
B=

B= 1.256 x 10^-25T  

     14.   A soft Iron ring has a mean diameter of 0.20m and an area of cross section 5×10^-4m² it is uniformly wound with 2000turns carrying a current of 2A and the magnetic  flux in the iron is 8x 10^-3Wb. What is the relative permeability of iron?

Solution

Length of ring l

l = 2

l = 2 x 0.10m

Number of turns per unit length n

n = =

If M is the absolute permeability of iron, then magnetic flux density of iron ring is

B =

B =

 Magnetic flux

 Magnetic flux  = BA

 Relative permeability of Iron μ_r

     15.   Two  flat  circular  coils  are  made  of  two  identical  wires  each of  length 20cm one  coil  has  number  of  turns  4  and  the  other  2. If the some   current flows though the wire in which will magnetic field at the centre will be greater?

Solution

 For the first coil

For second coil

 

 

 

 

 

 

 

 

 

   Therefore, magnetic field will be greater in coil with 4 turns   

     16.  A plat circular coil of 120 turns has a radius of 18cm and carries currents of 3A. What is the magnitude of magnetic field at a point on the axis of the coil at a distance from the centre equal to the radius of the coil?

Solution
Number of turns n = 120

Radius of the coil r = 0.18 m

Axial distance x = 0.18m

Current in coil I = 3A

 

                B = (4 x 10^-7) x 120 x3 x0.18²

                             2(0.18^{2 +} 0.18²) ^3/2

B= 4.4 x 10-4T

     17.  A current of 5A is flowing upward in a long vertical wire. This wire is placed in a uniform northward magnetic field of 0.02T. How much force and in which direction will this field exert on 0.06 length of the wire?

Solution

        
B = 0.02T

            I = 5A

            L = 0.06

               = 90^{0
F= 0.02 X 5 X 0.06Sin900
F = 0.006N}

 By Fleming’s Left hand rules the force is directed towards West

     18.   A straight wire of mass 200g and length 1.5m carries a current of 2A. It  is  suspend in  mind  air  by a  uniform  horizontal  magnetic  field  B.  What is the magnitude of the magnetic field?

solution

 M = 200 X 10^-3 kg

I = 2A

l = 1.5m

B =?
F=BIL

Mg = BIL

 B = Mg   = 200 x 10^-3 x 9.8

       IL                  1.5 X 2

 B = 0.65T

      19.    Two  long  horizontal  wires  are  kept  parallel  at a  distance of  0.2cm  apart  in a vertical plane . both the  wires  have  equal currents  in the  same  direction  the  lower  wire has  a  mass  of  0.05kg/m if the  lower  wire  appears weightless what  is  the  current  in  each  wire ?

Solution

Let  I  amperes be  the  current  in  each  wire the  lower  wire  is  acted upon by  two  forces.

Since the lower wire appears weightless the two forces were equal over 1m length of the wire

10^-4I² = 0.49

     20.   The  horizontal  component  of the  earth magnetic  field  at a  certain  place  is  3 x 10^-5 and  the  direction  of the  field  is  from  the  geographic  south  to the geographic  North  A very  long straight  conductor  is carrying  a steady current  of  1A. what  is the  force  per unit length on it when it  is  placed  on  a horizontal table  and  the  direction of  the  current  is

(a)   East  to  West

(b)   South to  North

Solution

(a)       When  current  is  flowing  from  east  to  west 90⁰

Force on the conductor per unit

 

 

(b)      When current  is  flowing from south to  north    =  0⁰  

    Force on the conductor per unit length

     21.  A horizontal  straight  wire 5cm long  of  mass 1.2gm^-1 placed perpendicular  to a  uniform magnetic  field  of  0.6T if  resistance  of the  wire  is  3.85cm^-1 calculate  the  P.d that  has  to be  applied  between  the  ends  of the  wire  to  make  it just  self supporting 

Solution

 The  current  (i) in the  wire  is to be in such  a direction  that  magnetic  force  acts on  it  vertically  upward. To  make  the  wire  self  supporting  its  weight  should be equal to the  upward  magnetic  force.

Resistance of the wire

               R = 0.05 x 3.8

                  = 0.19 

Required P. (I)      V = IR

                             V = 19.6X10^-3 X 0.19

                             V = 3.7 X 10^-3V

      22.   A  conductor  of  length  2m carrying  current  of  2A is  held  parallel to  an infinitely long  conductor carrying  current  of  10A  at  a  distance  of  100mm. find  the  force  on  small  conductor

Solution

 I_I  = 2A

I₂ = 10A

r = 100 x 10^-3m

l = 2m

Force  on  unit length  of  short  conductor  by the  long  conductor  is  give  by

   
Force  on  length  l  =  2m of  short  conductor  by the  long  conductor  is

     

   

  

   
The  force  will be  attractive  if  the  direction  of current  is the  same  in  two conduction  and  it will be  repulsive if  the  conductors  carry  current  in the  opposite  directions.

     23.   In the  figure  below, determine  the  position  between  two wire  which  experience  zero resultant force due to charge   Q  placed  at that  point

 

 

Solution
The  force  unit  length  acting  in  each wire  of the  parallel  wire  is  given by

Let    be  the  force  per unit  length in  the  wire  carrying  a  current  of  14A 

          Since  F₁ and  F₂ have  the  same  magnitude  but  they  are  acting in  opposite  direction for  resultant  force  to  be  zero

Assume  that  the  charge  Q is  placed  at  a distance  X from the  wire  carrying the

 

 

 

 

 

 

 

            The charge   Q  is placed 4m from the either wire.