11 A conductor when thermally heated possess large number offreely moving electron,even then no magnetic field is experienced near - Physics - Moving Charges And Magnetism

Question

11 A conductor when thermally heated possess large number
offreely moving electron,even then no magnetic field is
experienced near such a conductor What do you conclude?
12 A particle of mass m and charge q moves at right angles to
auniform magnetic field Plot a graph showing the variation of
theradius of the circular path described by it with the increase in
its
(a) charge,(b) kinetic energy,where ,in each case other
factorsremain constant Justify your answer
13 A charged paricle having a charge q ,is moving with a speed
valong the x-axis It enters a region of space where an electric
field E along y-axis and a magnetic field B are both present The
particle, on emerging from this region, is observed to be moving
along the x-axis only Obtain an expression for the magnitude of B
in terms of v and E Also give the direction of B
14 A long wire is first bent into a circular coil of one turn
and then intoa circular coil of smaller radius having n turns If
the same current passes in both the cases find the ratio of the
magnetic fieldproduced at the centers in the two cases
15 Why is diamagnetism independent of temperature?

Panchali Dutta · Accepted Answer

Answer to some of your questions:
11)When a conductor is heated the electrons start moving in
random direction, there is no one direction and thus there will be
no magnetic field around this conductor
12) The force experienced by the particle:
F = q(v X B) = qvB
mv2/r = qvB
r = mv/qB
Using the above expression, please try to plot the graph
a) radius is inversely proportional to the charge As the charge
increases the radius decreases
b) r is directly proportional to the energy So, if the particle
enters the magnetic field with greater energy it will describe a
bigger path
13) Lorentz force
F = qE +q(v X B)
v = v i
E = Ej
Since the particle emerges out of the region in its original
direction
FE = -FB
E = -vB
Magnitude of B = E/v
Direction of B is along the negative z axis

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