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F.Sc ICS Notes: Physics XII: Chapter 14 Electromagnetism Exercise Short Questions:

FSc ICS Notes: Physics XII: Chapter 14 Electromagnetism                                Exercise Short Questions:

Question 14.1 A plane conducting loop is located in a uniform magnetic field that is directed along the x-axis. For what orientation of the loop is the flux a maximum? For what orientation is the flux a minimum?
Answer 14.1 Magnetic flux will be maximum when the surface of the loop is perpendicular to the direction of magnetic field, the flux will be:
                                            φ = B • A
                                            φ = BAcosθ
                                            φ = BAcos0
                                            φ = BA (maximum)
Flux will be minimum when the surface of the loop is parallel to the direction of magnetic field and it is given by:
                                            φ = B • A
                                            φ = BAcosθ
                                            φ = BAcos90
                                            φ = 0 (minimum)                                          
which is the minimum value of flux.  

Question 14.2 A current in a conductor produces a magnetic field, which can be calculated using Ampere’s law. Since current is defined as the rate of flow of charge, what can you conclude about the magnetic field due to stationary charges? What about moving charges?
Answer 14.2 As we know that:
                                     (N∑ i=1) (B.ΔL)i = μo I
            According to Ampere's law we can write.
                                     (N∑ i=1) (B.ΔL)i = μo I
            For stationary charges      I = 0
                                      (N∑ i=1) (B.ΔL)i = 0
            If closed path is circular
                                      (N∑ i=1) B.ΔL =0
                                 (N∑ i=1) ΔL = 0
                       as          (N∑ i=1) ΔL = 2πr 0
                       so                    B = 0
            Hence magnetic field will not surround stationary charges.
            For moving charges I ≠ 0.
                                      B(2πr≠ 0    

Question 14.3 Describe the change in the magnetic field inside a solenoid carrying a steady current I, if (a) the length of the solenoid id doubled but the number of turns remains same and (b) the number of turns is doubled, but the length remains same.
Answer 14.3 Magnetic induction inside the solenoid is given by:
                              B = μo nI
            Where         n = N / l 
                              B = (μo NI) / l
            (a)    as length is doubled
                              B' = (μo NI) / 2l
                              B' = ½ (μo NI ) / l
                              B' = ½ B 
                     by making the length double field reduced to one half 
            (b)    as number of turns are doubled.
                              B' = (μo 2NI) / l
                              B' = 2 (μo NI ) / l
                              B' = 2 B
                       on doubling the number of turns magnetic field will be doubled.

Question 14.4 At a given instant, a proton moves in the positive x direction in a region where there is magnetic field in the negative z direction. What is the direction of the magnetic force? Will the proton continue to move in the positive x direction? Explain.
Answer 14.4  The force on the proton is given by:
                                   FB = +e (v x B)
              Where           v = vi
               and               B = B(-k)
                                   FB = +e (vi x B(-k))
                                   FB = -Bev (i x k)
                                   FB = -Bev (-j)
                                   FB = Bevj 
Therefore force on the proton is along positive Y axis. The proton will be deflected in the circular path.

Question 14.5 Two charged particles are projected into a region where there is a magnetic field perpendicular to their velocities. If the chargers are deflected in opposite directions,what can you say about them?
Answer 14.5 If two charged particles are projected into a region, where magnetic field is perpendicular to their velocities the charges are deflected in opposite directions. It means that particles are oppositely charged i.e. one is positively charged and other is negatively charged. If one is proton then other is electron.

Question 14.6 Suppose that a charge q is moving in a uniform magnetic field with a velocity v. Why is there no work done by the magnetic force that acts on the charge q?
Answer 14.6 The force on charge particle is always at perpendicular to its direction of motion. Thus force will deflect the charge particle in circular path. The work done by the force.
                                     W = F.d 
The angle between centripetal force and the displacement is 90 therefore the work done is given by:
                                     W = F.d Cos90
                                     W = 0 

Question 14.7 If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in the region is zero?
Answer 14.7 No we cannot say that the magnetic field in that region is zero because charge particle can move in a straight line when the force on it zero. This may be possible under two situations.
            (1) The direction of B and v are same i.e. θ = 0
                            We know that
                                                F = qvB Sinθ
                                                F = qvB Sin0
                                     or        F = 0
             (2) When v and B are in opposite direction i.e. θ = 180
                                               F = qvB Sinθ
                                               F = qvB Sin180
                                     or       F = 0

Question 14.8 Why does the picture on a TV screen become distorted when a magnet is brought near the screen?
Answer 14.8 The picture on T.V screen is formed by the electron beam. When a magnet is placed near the screen then the presence of the magnetic field of bar magnet the motion of charged particles is effected and their target on the screen of T.V is disturbed. Therefore, the image produced will be deformed.

Question 14.9 Is it possible to orient a current loop in a uniform magnetic field such that the loop will not tend to rotate? Explain.
Answer 14.9 The current carrying loop experience a torque when it is placed in the magnetic field. Which is given by:                 τ = IBAcosα
On orienting the loop, such that its plane is perpendicular to the magnetic field i.e.  α = 90, then
                              τ = IBAcos90
                              τ = 0 
Hence loop will not rotate when its plane is perpendicular to the magnetic field.

Question 14.10 How can a current loop be used to determine the presence of a magnetic field in a given region of space?
Answer 14.10 When a freely suspended loop of current experiences torque in the given region, then in that region magnetic field is present, otherwise not.

Question 14.11 How can you use a magnetic field to separate isotopes of chemical element? What should be the orientation of a current carrying coil in a magnetic field so that torque acting upon the coil is (a) maximum (b) minimum?
Answer 14.11 When Isotopes of the chemical element after ionization projected perpendicular to uniform magnetic field. they are deflected in circular path. The radius of circular path is given by:
                                           r = (mv) / (Bq)
                                           r m 
Therefore the heavier isotopes are deflected into circular path of larger radius as compared to lighter isotopes. Thus isotopes are differentiated on the basis of their radii in the magnetic field with the help of Mass spectrometer. 

Question 14.12 A loop of wire is suspended between the poles of a magnet with its plane parallel to the pole faces. What happened if a direct current is put through the coil? What happens if an alternating current is used instead?
Answer 14.12 We know that the torque experience by the current coil in magnetic field is given by:
                            τ = IBAcosα 
For maximum torque the plane of the coil is held parallel to the magnetic field i.e. α = 0
                            τ = IBAcos0 = BIA 
For minimum torque the plane of the coil is held perpendicular to the magnetic field i.e. α = 90
                            τ = IBAcos90 = 0

Question 14.13 Why the resistance of an ammeter should be very low?
Answer 14.13 As ammeter is always connected in series to measure the current, therefore it should have very small resistance. If an ammeter has high resistance then it will alter the current of the circuit.

Question 14.14 Why the voltmeter should have a very high resistance? 
Answer 14.14 The resistance of voltmeter should be very large so that it should not draw an appreciable amount of current on connecting it across a resistance to measure potential difference accurately.

Written By: Asad Hussain.