Q: A short bar magnet placed with its axis at 30^{o} with a uniform external magnetic field of 0.16\ T experiences a torque of magnitude 0.032 Nm If the bar magnet is free to rotate, its potential energies when it is in stable and unstable equilibrium are respectively

\theta =30^{o} B=0016\ T z=0.032 z=mB\sin\theta 0.032=m\times \dfrac{1}{2}\times 0.16 m=0.4\ Am U=-\vec{m}.\vec{B} U_{max}=mB = 0.4\times 0.16=0.064\ T U_{min}=-mB = -0.4\times 0.16= -0.064\ T

Q: A bar magnet is oscillating in the Earth's magnetic field with a period T. What happens to its period and motion if its mass is quadrupled

\text{Use the formula:}\ T=2\pi\sqrt{\dfrac{I}{MB}} \text{But:}\ I \propto m T^{'}=2\pi\sqrt{\dfrac{4I}{MB}} \Rightarrow \dfrac{T^{'}}{T}=2 T^{'}=2T

Q: The field intensity of two points on the axial line of a bar magnet at distances 40 cm and 20 cm is 10 : 125 . The length of the magnet is

\dfrac{\theta _{1}}{2}=tan^{-1}(\dfrac{l}{20}) , \dfrac{\theta _{2}}{2}=tan^{-1}(\dfrac{l}{40}) d_{1}=\sqrt{l^{2}+20^{2}} d_{2}=\sqrt{l^{2}+40^{2}} B_{3}=\dfrac{\mu _{0}p}{4\pi d_{2}^{2}} B_{4}=\dfrac{\mu _{0}p}{4\pi d_{2}^{2}} {B}''=\sqrt{B_{3}^{2}+B_{4}^{2}+2B_{3}B_{4}(cos\pi -\theta _{2})} cos\theta =\dfrac{1-tan^{2}\theta /2}{1+tan^{2}\theta /2} B_{1}=\dfrac{\mu _{0}p}{4\pi d_{1}^{2}} B_{1}=\dfrac{\mu _{0}p}{4\pi d_{1}^{2}} B_{2}=\dfrac{\mu _{0}p}{4\pi d_{1}^{2}} {B}'=\sqrt{B_{1}^{2}+B_{2}^{2}+2B_{1}B_{2}COS(\pi \theta _{1})} \dfrac{{B}'}{{B}''}=\dfrac{10}{125}=\dfrac{\dfrac{\mu _{0}p}{4\pi d_{1}^{2}}\sqrt{2(1+cos\theta _{1})}}{\dfrac{\mu _{0}p}{4\pi d_{2}^{2}}\sqrt{2(1-cos\theta _{2})}} \dfrac{2}{25}=\dfrac{(l^{2}+40^{2})\sqrt{1-(\dfrac{l}{20}p)/(l/20)^{2}}}{(l^{2}+20^{2})\sqrt{1-(1-(\dfrac{l}{40})^{2}}/(1+4d^{2})} l=20cm.

Q: The susceptibility of a magnetic material is -2.6 \times {10}^{-5} . Identify the type of magnetic material and state its two properties.

As the given susceptibility is small and negative so it will be diamagnetic material. The two properties of it are -i) When it is placed in uniform magnetic field, it is feebly magnetise opposite to the magnetic field.ii) The relative permeability is positive and less than one.

Q: (a) An iron ring of relative permeability \mu_{r} has windings of insulated copper wire of n turns per metre. When the current in the windings is I , find the expression for the magnetic field in the ring.(b) The susceptibility of a magnetic material is 0.9853 . Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.

(a) :Copper wire windings on the iron ring forms a toroid.Magnetic field due to a toroid is given by B = \mu n I where \mu is the permeability of the medium, n is the number of turns per unit length and I is the current flowing.Also, \mu = \mu_r \mu_o \implies \ B = \mu_r\mu_o nI (b) : Susceptibility of the magnetic material is given as \chi = 0.9853 Since the magnetic susceptibility of paramagnetic material is positive. So the given substance must be a paramagnetic material.

Question 1

A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth's magnetic field at the place is 0.39 G, and the angle of dip is  {35^ \circ }  .The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Accepted Answer

Number of horizontal wires in the telephone cable,  n=4 Current in each wire,  I=1.0 A Earth's magnetic field at a location , H=0.39 G=0.39  x   {10^{ - 4}} T Angle of dip at the location,  \delta  = {35^ \circ } Angle of declination, 	heta  \sim {0^ \circ } For a point 4 cm below the cable:Distance, r=4 cm=0.04 m The horizontal component of earth's magnetic field can be written as: {H_h} = H\cos \delta  - B Where ,B=Magnetic field at 4 cm due to current I in the four wires  = 4 	imes \frac{{{\mu _0}I}}{{2\pi r}} {\mu _0} =   Permeability of the free space   = 4\pi  	imes {10^{ - 7}}Tm\,{A^{ - 1}} 	herefore B = 4 	imes \frac{{4\pi  	imes {{10}^{ - 7}} 	imes 1}}{{2\pi  	imes 0.04}}  B = 2 	imes {10^{ - 9}}

Question 2

The angle of dip at a place is  40.6^{\circ}  and the intensity of the vertical component of the earth's magnetic field  V=6	imes 10^{-5}Tesla . The total intensity of the earth's magnetic field (l) at this place is

Accepted Answer

Given:-The angle of dip at a place is  40.6^o  intensity of the vertical component of the earth's magnetic field  V = 6 	imes 10^{&#8722;5}  \, Tesla From the relation    B_v = I sin \phi   I  = \dfrac {V}{sin \phi}= \dfrac { 6 	imes 10^{-5} }{ sin 40.6^0} = \dfrac { 6 	imes 10^{-5} }{ 0.65} = 9.2 	imes 10^{-5} tesla

Question 3

A short bar magnet placed with its axis at  30^{o}  with a uniform external magnetic field of  0.16\ T  experiences a torque of magnitude  0.032 Nm   If the bar magnet is free to rotate, its potential energies when it is in stable and unstable equilibrium are respectively

Accepted Answer

heta =30^{o} B=0016\ T z=0.032 z=mB\sin	heta  0.032=m	imes \dfrac{1}{2}	imes 0.16 m=0.4\ Am U=-\vec{m}.\vec{B} U_{max}=mB  =  0.4	imes 0.16=0.064\ T U_{min}=-mB  = -0.4	imes 0.16= -0.064\ T

Question 4

A bar magnet is oscillating in the Earth's magnetic field with a period T. What happens to its period and motion if its mass is quadrupled

Accepted Answer

ext{Use the formula:}\ T=2\pi\sqrt{\dfrac{I}{MB}} 	ext{But:}\ I \propto m T^{'}=2\pi\sqrt{\dfrac{4I}{MB}} \Rightarrow \dfrac{T^{'}}{T}=2 T^{'}=2T

Question 5

A solenoid has a core of a material with relative permeability  400 . The windings of the solenoid are insulated from the core and carry a current of  2A .If the number of turns is  1000  per meter, calculate (a)  H ,  (b)  B  and (c) the magnetising current  I_m .

Accepted Answer

Given :     \mu_r = 400         i = 2 A            n = 1000  turns per meter(a)    H = ni = 1000	imes 2 = 2000 A/m (b)    B = \mu_o H = 4\pi	imes 10^{-7}	imes 2000 = 8\pi	imes 10^{-4} \ T (c)    I_m = \chi H = 399	imes 8\pi	imes 10^{-4} = 10 \ A/m

Question 6

Two points A and B equidistant from the centre of a small bar magnet. Point &#8216;A&#8217; is on the axial line and point &#8216;B&#8217; is on the equatorial line. Then the magnetic fields at point &#8216;A&#8217; and &#8216;B&#8217; are in the ratio of

Accepted Answer

BA=\dfrac{2\mu _{0}m}{4\pi d^{3}}        B_{B}=\dfrac{\mu _{0}m}{4\pi d^{3}} \dfrac{B_{A}}{B_{B}}=2:1

Question 7

The field intensity of two points on the axial line of a bar magnet at distances  40  cm  and  20  cm  is  10 : 125 . The length of the magnet is

Accepted Answer

\dfrac{	heta _{1}}{2}=tan^{-1}(\dfrac{l}{20})   , \dfrac{	heta _{2}}{2}=tan^{-1}(\dfrac{l}{40}) d_{1}=\sqrt{l^{2}+20^{2}} d_{2}=\sqrt{l^{2}+40^{2}} B_{3}=\dfrac{\mu _{0}p}{4\pi d_{2}^{2}} B_{4}=\dfrac{\mu _{0}p}{4\pi d_{2}^{2}} {B}''=\sqrt{B_{3}^{2}+B_{4}^{2}+2B_{3}B_{4}(cos\pi -	heta _{2})} cos	heta =\dfrac{1-tan^{2}	heta /2}{1+tan^{2}	heta /2} B_{1}=\dfrac{\mu _{0}p}{4\pi d_{1}^{2}} B_{1}=\dfrac{\mu _{0}p}{4\pi d_{1}^{2}} B_{2}=\dfrac{\mu _{0}p}{4\pi d_{1}^{2}} {B}'=\sqrt{B_{1}^{2}+B_{2}^{2}+2B_{1}B_{2}COS(\pi 	heta _{1})} \dfrac{{B}'}{{B}''}=\dfrac{10}{125}=\dfrac{\dfrac{\mu _{0}p}{4\pi d_{1}^{2}}\sqrt{2(1+cos	heta _{1})}}{\dfrac{\mu _{0}p}{4\pi d_{2}^{2}}\sqrt{2(1-cos	heta _{2})}} \dfrac{2}{25}=\dfrac{(l^{2}+40^{2})\sqrt{1-(\dfrac{l}{20}p)/(l/20)^{2}}}{(l^{2}+20^{2})\sqrt{1-(1-(\dfrac{l}{40})^{2}}/(1+4d^{2})} l=20cm.

Question 8

The relative permeability of silicon is  0.99837  and that of palladium is  1.00692 , choose the correct option from the following :

Accepted Answer

Diamagnetic           Paramagnetic             Ferromagnetic-1  \leq   \chi  > 10  \leq   \mu_{r}  > 1 \chi   =  magnetic susceptibility \mu_{r}   =  relative permeability. \varepsilon  < 1 and small.

Question 9

The susceptibility of a magnetic material is  -2.6 	imes {10}^{-5} . Identify the type of magnetic material and state its two properties.

Accepted Answer

As the given susceptibility is small and negative so it will be diamagnetic material. The two properties of it are -i) When it is placed in uniform magnetic field, it is feebly magnetise opposite to the  magnetic field.ii) The relative permeability is positive and less than one.

Question 10

(a) An iron ring of relative permeability  \mu_{r}  has windings of insulated copper wire of  n  turns per metre. When the current in the windings is  I , find the expression for the magnetic field in the ring.(b) The susceptibility of a magnetic material is  0.9853 . Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.

Accepted Answer

(a) :Copper wire windings on the iron ring forms a toroid.Magnetic field due to a toroid is given by   B = \mu n I where  \mu  is the permeability of the medium,  n  is the number of turns per unit length and  I  is the current flowing.Also,   \mu = \mu_r \mu_o \implies \ B = \mu_r\mu_o nI (b) :  Susceptibility of the magnetic material is given as   \chi = 0.9853 Since the magnetic susceptibility of paramagnetic material is positive. So the given substance must be a paramagnetic material.

Question 11

Two short bar magnets of magnetic moments  0.125 Am^{2}  and  0.512 Am^{2}  are placed with their like poles facing each other. If the distance between the centres of the magnet is  0.26 m . The distance of neutral point from the weaker magnet is

Accepted Answer

B_{axial} = \dfrac{\mu_0 2M}{4\pi r^3} Equating the magnetic fields due to the two bar magnets,  \dfrac{M_1}{x^3} = \dfrac{M_2}{{(d-x)}^3} Solving the above for x (distance from the center of weaker magnet to the neutral point), gives the formula in the hint. Putting in the values, we get,  x = 0.1m

Question 12

A short bar Magnet with its north pole facing North forms a neutral point P in the horizontal plane. If the magnet is rotated by 90 ^0  in the horizontal plane, the net magnetic induction at P is:(Horizontal component of earths magnetic field is  B_H )

Accepted Answer

At P, the equatorial field due to magnet  =B_H .After rotation, P lies on the axis of the bar magnet.Hence, at P, field due to magnet  =2B_H Net field is the vector sum of  B_H  and  2B_H , both at right angles to each other giving,  B_{net} = \sqrt{5}B_H

Magnetism And Matter