Chapter 5 Magnetism And Matter Solutions
Question - 21 : - A compass needle free toturn in a horizontal plane is placed at the centre of circular coil of 30 turnsand radius 12 cm. The coil is in a vertical plane making an angle of 45ºwith the magnetic meridian. When the current in the coil is 0.35 A, the needlepoints west to east.
(a) Determine thehorizontal component of the earth’s magnetic field at the location.
(b) The current in thecoil is reversed, and the coil is rotated about its vertical axis by an angleof 90º in the anticlockwise sense looking from above. Predict the direction ofthe needle. Take the magnetic declination at the places to be zero.
Answer - 21 : -
Number of turns in thecircular coil, N = 30
Radius of the circularcoil, r = 12 cm = 0.12 m
Current in thecoil, I = 0.35 A
Angle of dip, δ =45°
(a) The magnetic fielddue to current I, at a distance r, is given as:Where,
= Permeability offree space = 4π × 10−7 Tm A−1
= 5.49 × 10−5 T
The compass needlepoints from West to East. Hence, the horizontal component of earth’s magneticfield is given as:
BH = Bsin δ
= 5.49 × 10−5 sin45° = 3.88 × 10−5 T = 0.388 G
(b) When the current inthe coil is reversed and the coil is rotated about its vertical axis by anangle of 90 º, the needle will reverse its original direction. In this case,the needle will point from East to West.
A magnetic dipole is under theinfluence of two magnetic fields. The angle between the field directions is60º, and one of the fields has a magnitude of 1.2 × 10−2 T. Ifthe dipole comes to stable equilibrium at an angle of 15º with this field, whatis the magnitude of the other field?
Answer - 22 : -
Magnitude of one of the magneticfields, B1 = 1.2 × 10−2 T
Magnitude of the other magneticfield = B2
Angle between the twofields, θ = 60°
At stable equilibrium, the anglebetween the dipole and field B1, θ1 =15°
Angle between the dipole andfield B2, θ2 = θ − θ1 =60° − 15° = 45°
At rotational equilibrium, thetorques between both the fields must balance each other.
∴Torque due to field B1 = Torque due tofield B2
MB1 sinθ1 = MB2 sinθ2
Where,
M =Magnetic moment of the dipole
Hence, the magnitude of the othermagnetic field is 4.39 × 10−3 T.
Question - 23 : - A monoenergetic (18 keV) electronbeam initially in the horizontal direction is subjected to a horizontalmagnetic field of 0.04 G normal to the initial direction. Estimate the up ordown deflection of the beam over a distance of 30 cm (me=9.11 × 10−19 C). [Note: Data inthis exercise are so chosen that the answer will give you an idea of the effectof earth’s magnetic field on the motion of the electron beam from the electrongun to the screen in a TV set.]
Answer - 23 : -
Energy of an electron beam, E =18 keV = 18 × 103 eV
Charge on an electron, e =1.6 × 10−19 C
E =18 × 103 × 1.6 × 10−19 J
Magnetic field, B =0.04 G
Mass of an electron, me =9.11 × 10−19 kg
Distance up to which the electronbeam travels, d = 30 cm = 0.3 m
We can write the kinetic energyof the electron beam as:
The electron beam deflects alonga circular path of radius, r.
The force due to the magnetic fieldbalances the centripetal force of the path.
Let the up and down deflection of the electron beambe 
Where,
θ =Angle of declination
Therefore, the up and downdeflection of the beam is 3.9 mm.