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151.

A narrow beam of identical ions with specific charge `q//m`, possessing different velocities, enters the region of space, where there are unifrom parallel electric and marnetic fields the strength `E` and induction `B`, at the point `O` (see Fig). The beam direction coincides with the `x` axis at the point `O`. A plane screen oriented at right angles to the `x` axis is located at a distance `l` from the point `O`. Find the equation of the trace that the ions leave on the screen. Demonstrate that at `x lt lt l` it is the equaction of a parabola.

Answer» The equcation of the trajectory is,
`x = (v_(0))/(omega) sin omega t, z = (v_(0))/(omega) (1 - cos omega t), y = (qE)/(2m) t^(2)` as before see (3.384)
Now on the screen `x = l`, so
`sin omega t = (omega l)/(v_(0))` or, `omega t = sin^(-1) (omega l)/(v_(0))`
At that moment,
`y = (qE)/(2m omega^(2)) ("sin"^(-1)(omega l)/(v_(0)))^(2)`
so, `(omega l)/(v_(0)) = sin sqrt((2m omega^(2) y)/(qE)) = sin sqrt((2q B^(2) y)/(Em))`
and `z = (v_(0))/(omega) 2 "sin"^(2) (omega t)/(2) = l "tan" (omega t)/(2)`
`= l "tan" (1)/(2) ["sin"^(-1) (omega I)/(v_(0))] = l "tan" sqrt((qB^(2) y)/(2 mE))`
For small
`z, (qB^(2) y)/(2 mE) = ("tan"^(-1) (z)/(l))^(2) = (z^(2))/(l^(2))`
or, `y = (2 mE)/(q B^(2) l^(2)). z^(2)` is a parabola.
Discussion
152.

In a certain region of the inertial reference frame there is magnetic field with induction `B` rotating with angluar velocity `omega`. Find `grad xx E` in this region as a function of vectors `omega` and `B`

Answer» A rotating magnetic field can be represented by,
`B_(x) = B_(0) cos omega t, B_(y) = B_(0) sin omega t` and `B_(x) = B_(zo)`
Then curl, `vec(E) = - (del vec(B))/(del t)`,
So, `-(Curl vec(E))_(x) = -omega B_(0) sin omega t = -omega B_(y)`
`-(Curl vec(E))_(y) = omega B_(0) cos omega t = omega B_(x)`, and `-(Curl vec(E))_(x) = 0`
Hence, Curl `vec(E) = -vec(omega) xx vec(B)`,
where, `vec(omega) = vec(e_(3)) omega`.
Discussion
153.

A long solid aluminum cylinder of radius `a = 5.0 cm` rotates about its axis in unidrom magnetic field with induction `B = 10 mT`. The angluar velocity of rotation equlas `omega = 45 rad//s` with `omega uarr uarr B` Neglecting the magnetic field of appearing chagres, find their spcae and surfaface densities.

Answer» Choose `vec(omega) uarr uarr vec(B)` along the z-axis and choose `vec(r)`, as the cylindrical polar radius of a reference point (perpendicular distance from the axis). This point has the velocity.
`vec(v) = vec(omega) xx vec(r)`,
and experiences a `(vec(v) xx vec(B))` force, which must be counterbalanced by an electric field,
`vec(E) = -(vec(omega) xx vec(r)) xx vec(B) = -(vec(omega). vec(B)) vec(r)`.
There must appear a space charge density,
`rho = epsilon_(0) div vec(E) = -3 epsilon_(0) vec(omega) vec(B) = -8 pC//m^(3)`
Since the cylinder, as a whole is electrically neutral the surface of the cylinder must acquire a positive charge of surface density,
`sigma = + (2 epsilon_(0) (vec(omega). vec(B)) pi a^(2))/(2pi a) = epsilon_(0) a vec(omega).vec(B) = +- 2 pC//m^(2)`
Discussion
154.

A large plate of non-ferromagnetic material moves with a constant velocity `v = 90 cm//s` in a unifrom magnetic field with induction `B = 50 mT` as shown in Fig. Find the surface density of electric charges appearing on the plate as a result of its motion.

Answer» Within the plate, there will appear a `(vec(v) xx vec(B))` force, which will cause charges inside the plate to drift, until a countervalling electric field is set up. This electric field is related to `B`, by`E = e B` since `v & B` are mutually perpendicular, and `E` is perpendicular to both. THe charge density `+- sigma`, on the force of the plate, producing this electric field, is given by
`E = (sigma)/(epsilon_(0))` or `sigma = epsilon_(0) v B = 0.40 pC//m^(2)`
Discussion
155.

A charged particle moves along a circle of radius `r = 100 mm` in a unifrom magnetic field with induction `B = 10.0 mT`. Find its velocity and perios of revolution if that particle is (a) a non-relativistic proton, (b) a relativistic electron.

Answer» (a) For motion along a circle, the magnetic froce, acted on the particle, will provide the centripetal force, neccesary for its circular motion.
`i.e. (mv^(2))/(R) = evB` or, `v = (eBR)/(m)`
and the period of revolutin, `T = (2pi)/(omega) = (2piR)/(v) = (2pi m)/(eB)`
(b) Generally, `(d vec(p))/(dt) = vec(F)`
But, `(d vec(p))/(dt) = (d)/(dt) (m_(0) vec(v))/(sqrt(1 - (v^(2)//c^(2)))) = (m_(0) dotvec(v))/(sqrt(1 - (v^(2)//c^(2)))) + (m_(0))/((1 - (v^(2)//c^(2)))^(3//2)) (vec(v) (vec(v) . dotvec(v)))/(c^(2))`
For transerverse motion, `vec(v). dotvec(v) = 0` so,
`(d vec(p))/(dt) = (m_(0) dotvec(v))/(sqrt(1 - (v^(2)//c^(2)))) = (m_(0))/(sqrt(1 - (v^(2)//c^(2)))) (v^(2))/(r)`, here
Thus, `(m_(0) v^(2))/(r sqrt(1 - (v^(2)//c^(2)))) = B ev` or, `(v//c)/(sqrt(1 - (v^(2)//c^(2)))) = (B er)/(m_(0) C)`
or, `(v)/(c) = (B er)/(sqrt(B^(2) e^(2) r^(2) + m_(0)^(2) c^(2)))`
Finally, `T = (2pi r)/(v) = (2pi m_(0))/(eB sqrt(1 - v^(2)//c^(2))) = (2pi)/(cBe) sqrt(B^(2) e^(2) r^(2) + m_(0)^(2) c^(2))`
Discussion
156.

The magnetic field B inside a long solenoid, carrying a current of 5.00 A, is `(3.14 xx 10^-2)` T. Find the number of turns per unit length of the solenoid.

Answer» Correct Answer - B
Discussion
157.

A very long straight solenoid has a cross section radis `R` and `n` turns per unit lenghth. A direct current `I `flows throguh the solenoid. Suppose that `x` is the distance from the end of the the solenoid, measured along its axis. Find: (a) the magnetic induction `B` on the axis as a funciton of `x`, draw an approximate plot of `B` vs ratio `x//R`, (b) the distance `x_(0)` to the point on the axis at which the value of `B` differs by `eta = 1%` from that in the middle section of the solnoid.

Answer» We proceed exactly as in the previious problem. Then (a) the magnitude induction on the axis at a distance `x` from one end is clearly.
`B = (mu_(0) nI)/(4pi) xx 2pi R^(2) int_(0)^(oo) (dz)/([R^(2) + (z -x)^(2)]^(3//2)) = (1)/(2) mu_(0) n I R^(2) int_(x)^(oo) (dz)/((z^(2) + R^(2))^(3//2))`
`= (1)/(2) mu_(0) n I int_("tan"^(-1) (x)/(R))^(pi//2) cos theta d theta = (1)/(2) mu_(0) n I (1 - (x)/(sqrt(x^(2) + R^(2))))`
`x gt 0` means that the field point is outside the solenoid. B then falls with `x. x lt 0` means that the field point gets more and more inside the solenoid . `B` then increases wtih`(x)` and eventually becomes constant equal to `mu_(0) n I`. The `B - x` graph is as given iin the answer script.
(b) We have, `(B_(0) - del B)/(B_(0)) = (1)/(2) [1 - (x_(0))/(sqrt(R^(2) + x_(0)^(2)))] = 1 - eta`
or, `- (x_(0))/(sqrt(R^(2) + x_(0)^(2))) = 1 - 2eta`
Since `eta` is small `(oo 1%), x_(0)` must be negative. Thus `x_(0) = -|x_(0)|`
and `(|x_(0)|)/(sqrt(R^(2) + |x_(0)|^(2))) = 1 - 2eta`
`|x_(0)|^(2) = (1 - 4eta + 4eta^(2)) (R^(2) + |x_(0)|^(2))`
`0 = (1 - 2 eta)^(2) R^(2) - 4eta (1 - eta) |x_(0)|^(2)`
or, `|x_(0)| = ((1 -2 eta) R)/(2sqrt(eta (1 - eta)))`
Discussion
158.

Calculate the inducatance of a doughunt solenoid whose inside radius is equal to `b` and cross-section has the form of a square with side `a`. The solenoid winding consists of `N` turns. The space inside the solenoid is filled up with unifrom paramagnetic having permeability `mu`.

Answer» Within the solenoid, `H_(varphi) 2pi r = N I` or `H_(varphi) = (NI)/(2pi r), B_(varphi) = mu mu_(0) (N I)/(2pi r)`
and the flux, `Phi = N Phi_(1) = N (mu mu_(0))/(2pi) In int_(b)^(a + b) (a dr)/(r)`
Finally, `L = (mu mu_(0))/(2pi) N^(2) a In (1 + (a)/(b))`
Discussion
159.

Demonstarate that the magnetic energy of interaction of two current-carrying loops located in vacumm can be represented as `W_(ia) = (1/mu_(0)) int B_(1) B_(2) dV`, where `B_(1)` and `B_(2)` are the magnetic inductions within a volume element `dV`, produced indiviually by the currents of the first and the secound loop respectively.

Answer» The interaction energy is
`(1)/(2 mu_(0)) int |vec(B_(1)) + vec(B_(1))|^(2) dV - (1)/(2 mu_(0)) int |vec(B_(1))|^(2) dV - (1)/(2 mu_(0)) int |vec(B_(2))|^(2) dV`
`= (1)/(mu_(0)) int vec(B_(1)). vec(B_(2)) dV`
Here, if `vec(B_(1))` is the magnetic field produced by the first of the current carrying loops, and `vec(B_(2))` that of the secound one, then the magnetic field due to both the loops will `vec(B_(1)) + vec(B_(2))`.
Discussion
160.

Find the interaction energy of two loops carrying currents `I_(1)` and `I_(2)` if both loops are shaped as circles of radii `a` and `b`, with `a lt lt b`. The loops centres are located at the same point and their planes from an angle `theta` between them.

Answer» We can think of the smaller coil as consituting a magnet of dipole moment,
`p_(m) = pi a^(2) I_(1)`
Its direction is normal to the loop and makes and angle `theta` with the direction of the magnetic field, due to the bigger loop. This magnetic field is,
`B_(2) = (mu_(0) L_(2))/(2b)`
The interactiion energy has the magnitude,
`|W| = (mu_(0) I_(1) I_(2))/(2b) pi a^(2) cos theta`
Its sign depends on the sense of the currents.
Discussion
161.

A small coll is intorduced between the poles of an electromagnent so that its axis coincides with the magnetic field direction. The cross sectional ara of the coil is equal to `S = 30 mm^(2)`, the its diameter a balistic galvanometer connected to the coil indicates a charge `q = 4.5 muC` flowing through it. Find the magnetic induction magnitude between the poles provided the total resistance of the electric circuit equals `R = 40 Omega`

Answer» The flux through the coil changes sign. Initirally it is `BS` per turn.
Finally it is `-BS` per turn. Now if flux is `Phi` at an intermediate state then the current at that moment will be
`I = (-N (d Phi)/(dt))/(R)`
so charge that flows during a sudden turing of the coil is
`q = int i dt = -(N)/(R)[Phi -(-Phi)] = 2 N BS//R`
Hence, `B = (1)/(2) (q R)/(N S) = 0.5 T` on putting the values.
Discussion
162.

A `pi` shaped metal frame is located in a uniform magnetic field perpendicular to the plane of the conductor and varying with time at the rate `(dB//dt)=0.I0T//sec` . A conducting connector starts moving with an acceleration `a=I0cm//sec^(2)` along the parallel bars of the frame. The lenght o0f the connector is equal to `l=20cm` . Find the emf induced in the loop `t=2sec` after the beginnig of the motion, if at the moment `t=0` the loop area and the magnetic induction are equal to zero. The inductance of the loop is to be neglected.

Answer» The flux through the loop changes due to the variarion in `vec(B)` with time and also due to the movement of the connector.
So, `xi_(In) = |(d(vec(B).vec(S)))/(dt)| = |(d(BS))/(dt)|` as `vec(S)` and `vec(B)` are colliniear
But, `B`, after `t` sec of begining of motion = `Bt`, and `S` becomes `= l (1)/(2) w t^(2)`, as connector starts moving from rest with a constant accelertion `w`.
So, `xi_("ind") = (3)/(2) B l w t^(2)`
Discussion
163.

A slightly divergent beam of non-relatistic charged particles accelearated by a potential difference `V` propagates from a point `A` along the axis of a straight solenoid. The beacm is brought into focus at a distance `l` from the point `A` at two successive values of magnetic induction `B_(1)` and `B_(2)`. Find the spefic charge `q//m` of the particles.

Answer» The charged particles will traverse a helical trajectroy and will be focussed on the axis after traversing a number of turns. Thus
`(l)/(v_(0)) = pi. (2pi m)/(q B_(1)) = (pi + 1) (2pi m)/(q B_(2))`
So, `(n)/(B_(1)) = (pi + 1)/(B_(2)) = (1)/(B_(2) - B_(1))`
Hence, `(l)/(v_(0)) = (2pi m)/(q(B_(2) - B_(1)))`
or, `(l^(2))/(2qV//m) = ((2pi)^(2))/((B_(2) - B_(1))^(2)) xx (1)/((q//m)^(2))`
or, `(q)/(m) = (8 pi^(2) V)/(l^(2) (B_(2) - B_(1))^(2))`
Discussion
164.

Magentron is a device consisting of a filament of radius `a` and coaxial cylindrical anode of radius `b` which are located in a unifrom magnetic field parallel to the filament. An accelerating potential differnece `V` is applied between the filament and the anode. Find the value of magnetic induction at which the electrons leaving the filamnent with zero velocity reach the anode.

Answer» This differs from the previous problem in `(a harr b)` and the magnetic field is along the z-direction. Thus `B_(x) = B_(y) = 0, B_(x) = B`
Assuming as usual the charge of the electron to be `-|e|`, we write the equaction of motion
`(d)/(dt) mv_(x) = (|e| V_(x))/(rho^(2) In (b)/(a)) = -|e| B doty, (d)/(dt) mv_(y) = (|e| V_(y))/(rho^(2) In (b)/(a)) + |e| B dotx`
and `(d)/(dt) mv_(x) = 0 implies z = 0`
The motion is confined to the plane `z = 0`. Eliminating `B` from the first two equactions,
`(d)/(dt) ((1)/(2) mv^(2)) = (|e| V)/(In b//a) (x dotx + y doty)/(rho^(2))`
or, `(1)/(2) mv^(2) = |e| V (In rho//a)/(In b//a)`
so, as expected since magnetic forces do not work,
`v = sqrt((2|e|V)/(m))`, at `rho = b`.
On the other hand, elaminating `V`, we also get,
`(d)/(dt)m (xy_(y) - yv_(x)) = |e| B (x dotx + y doty)`
`i.e. (xy_(y) - yv_(x)) = (|e| B)/(2m) rho^(2) +` constatn
The constant is easily evaluated, since `v` is zero at `rho = a`. Thus,
`(xy_(y) - yv_(x)) = (|e| B)/(2m) (rho^(2) - a^(2)) gt 0`
At `rho = b, (xv_(y) - yv_(x)) le vb`
Thus, `vb ge (|e|B)/(2m) (b^(2) - a^(2))`
or, `B le (2 mb)/(b^(2) - a^(2)) sqrt((2 |e|V)/(m)) xx (1)/(|e|)`
or, `B le (2b)/(b^(2) - a^(2)) sqrt((2mB)/(|e|))`
Discussion