Physicsmynd

Given below is a last minute revisioner of Magnetism for the IIT JEE Physics paper . Note that the points covered are based on the exam point of view , and is not meant to be a comprehensive review of the entire magnetism portion.

• A moving charge / flowing electric current produces a magnetic feild ( which is a vector feild ) in addition to the electric force / feild . If a charged particle moves parallel to the magnetic feild , the magnetic force on the charge is zero . The magnetic force differs from an electric force in the fact that a) it depends on the particle velocity b) it’s direction is perpendicular to both |v| and |B| and c) it does no work in displacing a charged particle , provided the magnetic feild is constant.
• Consider a charged particle of mass m moving in an uniform magnetic feild | B | having an initial velocity vector | v | perpendicular to the feild. Then , the path radius = r = mv / qB , angular speed ω = qB/m and time period, T = 2 π m/qB.If the motion in the uniform magnetic feild is at some arbitrary angle θ with respect to B , then the path becomes helical – here there are two velocity components, the perpendicular component moving the charge in a circular path of radius given by r = mv1/qB and the parallel component moving along the feild lines. Pitch p = 2 π m vΦ/qB.
• Consider a charged particle moving in an electric feild which is perpendicular to the magnetic feild . Here , the particle velocity is perpendicular to both the feilds. The magnetic feild will rotate the particle in a circle in the x-z plane , || to the magnetic feild resulting in a helical path with increasing pitch . vx = vo cos ( Bqt/m ) and vz = vo sin ( Bqt/m ) .
• A current carrying conductor of any arbitrary shape in an uniform magnetic feild will experience a force |F| = I|L|x|B| where |L| is the length vector joining initial and final points of a conductor. If it’s a closed loop then  |F| = zero. If the magnetic feild is not uniform , various elements of the loop will experience different forces - means that if the loop is free enough , the loop could have a circular motion , ie , |Fr| = zero, but |ζ | may or may not be zero.
• Considering the force between two infinite , parallel current carrying conductors , the force / unit length = μo I I’/2 π r . Note that the if the direction of current flow is the same , the wires will attract each other. 
• As the magnetic feild pattern produced by a small current loop is like a bar magnet, it acts like a magnetic dipole . Here , if the loop is not lying in a single plane , two equal and opposite currents are to be assumed in a single branch [ note that the net change is zero ] and likewise the required loops are to be considered in different planes . For eg , for a cube of side I carrying a current i  , |M| net = – i I² ( iˆ + kˆ ) .
• If a magnetic dipole changes it’s orientation in a magnetic feild by an infinitesimal angular displacement d Φ , the feild does work dW given by ζ d Φ  =  dU , the change in potential energy . The energy present in the loop is U = – |M| |B| .
• A moving point charge with velocity v creates circular magnetic feilds centered on the line of v and which are perpendicular to it. The magnetic feild , |B| = ( μo /4π ) q|v|rˆ/r² .
• For a current carrying straight wire , the feild is zero for points along the length of the wire , but not on it. Here , as the feild is perpendicular to the plane containing both the wire and the point , the lines of force are concentric circles which encircle the wire. For the infinite long wire , if the point is near the end , B = ( μo /4π ) 2I/d and if the point is near one end , then, B = ( μo /4π ) I/d .
• The magnetic feild at the centre of a current carrying arc is B = μo I/2 R  b)  a point inside a long solenoid B = μo n I  c) for a point at one end of a long solenoid  B =  μo n I /2  and d) for a point at a distance R from the centre of a flat strip of width a along it’s perpendicular bisector, B = ( μo i /π a ) tan -¹ ( a/2R )  Note – in cases where the points considered are far from the strip , B = ( μo /2π ) ( i/R ) .

The last minute revisioner given below covers the salient points of the relevant sections from the exam point of view . It’s not meant to be a comprehensive review of the entire chapters. 

• Charge is an intrinsic ( it’s own ) , individualistic and indestrucible property of a body/particle which results in electric forces.  No of charges can be reduced or increased by friction , conduction or induction. Charges are quantized , conserved and additive. A non-accelerated moving charge also causes magnetic forces in addition to the electric forces. If it’s accelerated , the charge also radiates energy.
• The dielectric constant ( K = ε_r = ε /εo)  in Couloumb’s law is usually ≥ 1 . When a charge is placed in a medium , the force decreases K times. Coulomb’s law a) applies only to point charges b) obeys Newton’s third law c) is not applicable to distances below 10 ^-15  [ - fifteen ] m and d) follows the superposition principle. Note that the interaction between two charges is not influenced by a third charge . 
• The period of revolution of a particle of mass m of charge – q1 moving around a fixed charge + q2 in a circular path of radius r is given by T = √  ( 16 π³ εo m r³ / q1 q2 )
• Consider a thin fixed ring of radius a having a positive charge q uniformely distributed over it . Now , if a particle of mass m and having a negative charge Q is placed on the axis at a distance of x ( x << a ) from the ring’s centre , the negative charge  undergos oscillatory motion nearing simple harmonic as the restoring force is not linear. T = √ 2 π ( 4 π εo m a³ /q Q )
• To keep the original charge distribution unchanged , test charges are usually of a small value. If a test charge doesn’t experience a force at a point  , the net electric feild at that point should be zero. Note that a charged particle is not affected by it’s own feild.
• Electric feild lines helps us to visualize the nature of electric feild in a given space. They a) never cross each other b) never form closed loops and c) doesn’t pass through a conductor. Eventhough the tangent to the line of force in an electric feild at a point gives the direction of force / acceleration  acting on a positive charge at that point , it isn’t necessary that the charge will move in the said direction of force. 
• The maximum value of electric feild of a ring shaped conductor having radius r ,carrying a total uniformely distributed charge Q at a point P which lies on the axis of the ring at a distance x from it’s center is 1/4 π εo ( 2Q /[ 3 √ 3 R² ] )
• The electric feild produced by an infinite plane sheet of charge is independent of the distance from the sheet . This feild is uniform and it’s direction is perpendicular to the sheet and away from it.
• The net electric feild of a linear charge / charged ring / semicircular charge and a circular charged arc is ∫ d E cos θ . The net electric feild of a charged rod of fixed length having charged density λ along x axis is λ / 4 π εo ro ( sin θ1 + sin θ2 ) and along y axis is λ / 4 π εo ro ( cos θ1 – cos θ2 )  , of a  semicircular ring having charge density λ - x axis λ / 2 π εo r and along y axis is zero , of a quarter circular ring of charge density λ – x axis and y axis  is λ / 4 π εo r.
• Electric feild intensity of a short dipole at some general point is [ p / 4 π εo r³ ] √ (1 + 3 cos² θ ) . Here the resultant feild intensity vectors E2 / E1 = tan Φ
• Electric flux through a surface describes whether electric feild points into or out of the surface. In an uniform feild , the flux through a closed surface is zero. For a surface of area A tilted at an angle Φ , flux is E A cos Φ and volume flow rate through A is dV/ dt = v A cos Φ .
• Gauss’s  law is most useful in dealing with charge distributions of spherical / cylindrical symmentry or when charge distribution is uniform over a plane. For point / spherical charge distributions gaussian surface is spherical while that for a line of charge it’s cylindrical. For feild of a line charge it’s 1 / 2 π εo ( λ/r ) , for an infinite plane sheet of charge  σ / 2 εo  and at the surface of a conductor it’s σ / εo , inside a uniformely charged sphere ρ r / 3 εo  and outside is  ( 1/4π εo ) ( Q / r² ) . Electric feild of a long uniformely charged cylinder is ρ R² / 2 εo r and that inside a uniform volume charged plane is  ρ r/εo and that of the outside is ( ρ /εo ) d/2
• The work done by an electric force to move a point charge through a small displacement dl depends only on the change in the distance dr between the charges ( i.e , the end points are important and not the path taken ), which is the radial component of the displacement. Hence the force acting is conservative. The potential energy U = 1/4π εo (q qo/r ) .
• Electric potential is a scalar quantity where the principle of superposition can be applied . This means that for a discrete distribution of charges where V = V1 + V2 + V3 +… , = ( 1/4π εo ) Σ q/r .In case the potential is caused by a continuous charge distribution , then , V = ∫ dq /4π εo .
• Equipotential surfaces can be volumes , surfaces or lines . Note – a) they will never cross each other as the potential at a point can’t have two values b) for rest charges , the conducting surfaces are always equipotential c) equipotential surfaces are always perpendicular to the electric lines of force and d) the work done in moving a charge from a point to another point in an equipotential surface is zero.
• The cartesian components of electric feild w.r.t the electric potential are Ex = – ∂V/∂x , Ey = – ∂V/∂y and Ez = – ∂V/∂z  . Note – the equation indicates that the negative of the rate of change of potential with position in any direction is the component of E in that direction. Negative sign means that E is pointing in the direction of decreasing V.
• For an infinite rod , the potential at any point near the rod is infinite . But the potential difference between any two points a and b is  Va -Vb  =  ( λ / 2 π εo  ) In ( b/a ) .