Q.220. Motion of proton in a magnetic field.

Question 220.
A proton is projected with a speed of 500 m/s in a uniform magnetic field of 5T. If the angle between velocity & force vector is 45°.  Find the pitch of its helical path.
 Answer 220.
Speed parallel and perpendicular to the magnetic field
= 500cos(45°)
= 250sqrt(2) m/s.
Radius of the helical curve is given by
qvB = mv^2/r
=> r = mv/qB
=> r
= (1.67262158 × 10-27)*(250sqrt(2)) / (1.6x10^-19 * 5) m
= 7.392 x 10^-8 m

 Pitch of the helical path of the proton,
= 2πr/(vcos45°) * vsin45
= 2πr
= 2 x 3.14159 x 7.392 x 10^-8 m
= 4.64 x 10^-7 m.

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Q.193. Motion of a charged particle in an electromagnetic field.

Question 193.
A velocity selector uses a 6o mT magnetic field perpendicular to 24 kN/C electric field. At what speed will a proton pass through the selector undeflected?
(a) After emerging from the velocity selector, the proton enters a mass spectrometer with a 1 T magnetic field perpendicular to the velocity. What is the diameter x of the semicircular trajectory of the proton just before it impinges on the detector?
(b) What would be deflection x be if the charged particle is an electron going through the same velocity selector and the mass spectrometer as above?

Answer 193.
v = E/B = 24000/0.06 = x * 10^5 m/s.

(a)
Force on proton of mass m moving on trajectory of radius r equals the force due to magnetic field
=> mv^2/r = qvB
=> diameter = 2r
= 2mv/qB ... [m = mass of proton and q = charge of proton]
= 2 (1.66 x 10^-27) (4 x 10^5) / (1.6 x 10^-19 x 1)
= 8.3 x 10^-3 m.

(b)
The force on the electron due to electric field and magnetic field will still be equal but opposite in direction to that on proton. Hence, electron will also move undeflected.
diameter of the trajectory of electron
= 8.3 x 10^-3 / 1836
... [because mass of electron is 1/1836 of the mass of proton and charge is the same as that of proton]
= 4.52 x 10^-6 m.

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Q.188. Faraday's law of electromagnetic induction

Question 188.                                                       

Figure shows a bar of mass m = 0.2 kg

that can slide without friction on a pair

of rails separated by a distance l =1.2 m

and located on an incline plane that

makes an angle = 25.0° with respect to

the ground. The resistance of the resistor

is R = 1.0 and a uniform magnetic field

of magnitude of B = 0.5 T is directed

downward, perpendicular to the ground,

over the entire region through which the

bar moves. With what speed v

does the bar slide along the rails?

Answer 188.

Component of B perpendicular to the plane
= Bcos25°

=> emf induced in the coil
ε = Bcos25° * 1.2 * v =
=> current in the bar,
I = ε/R = (v/R) * 1.2 Bcos25°
Force opposing the downward motion
F = I * L * Bcos25° = (v/R) * (1.2 Bcos25°)^2

As the bar starts sliding, v keeps on increasing and so does F till F balances the downward component of the weight of the bar
=> mgsin25° = (v/R) * (1.2 Bcos25°)^2
=> velocity, v
= (mgRsin25°) / (1.2 Bcos25°)^2
= (0.2 * 9.81 * 1.0 * sin25°) / (1.2 * 0.5 * cos25°)^2
= (0.8292) / (0.2957)
= 2.80 m/s.

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Q.78. Electromagnetic Induction.

Question 78.

A conducting rod of length l is moved at constant velocity Vo on two parallel, conducting, smooth, fixed rails, that are placed in a uniform constant Magnetic field B perpendicular to the plane of rails as shown in the figure. A resistance R is connected between the two ends of the rail, then please explain how the following options are correct?

1) Thermal power dissipated in the resistor is equal to the rate of work done by the external person pulling the rod.

2) If applied external force is doubled, then a part of the external power increases the velocity of the rod.

3) If resistance R is doubled, then power required to maintain the constant velocity v becomes half.

Answer 78.

1)

If "I" is the current flowing through the conducting rod and its length is "L", then the force acting on it due to magnetic field in the loop of intensity B is
F = IBL
=> mechanical power developed ,
P = FVo = IBLVo ... ( 1 )

Thermal power dissipated
= electric power expended in the wire
= emf developed x current
= BLVo * I
= IBLVo ... ( 2 )

From ( 1 ) and ( 2 ),
thermal power dissipated in the resistor
= work done by the external force.

2)
When force F is acting on the conducting rod, velocity is constant at Vo and according to Newton's first law, no net force is working on the rod. This is because, the force applied balances the force developed on the conducting rod in the opposite direction due to rate of change of magnetic flux. Now when F is increased to 2F, initially net force is 2F - F = F which accelerates the rod and increases its velocity due to the acceleration. With increase in velocity, the opposite magnetic force keeps increasing and when the velocity doubles to Vo becomes equal to 2F. At that instant, again the net force becomes zero and the rod continues to move with velocity 2Vo. During this period, net force changes from F to zero, and the varying net force increases the velocity of the rod from Vo to 2Vo.

3)
If R is doubled, current I becomes half and the force required to maintain velocity also becomes half. If a constant velocity Vo is maintained in this condition, then the power required also becomes half as the force is half and velocity is the same since power = force x velocity.

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Q.30. Electromagnetism

Question 30.
A particle having mass m and charge q is released from the origin in a region in which magnetic field and electric field are given by B vector = - Bo*j vector and E vector = Eo* k vector. Find the speed of the particle as a function of its Z - coordinate.

Answer 30.
Rozeta53, one of my most valuable contacts in Yahoo Answers, provided the complete solution to this problem which is reproduced hereunder with thanks.

Lorentz force: F = (Fx, Fy, Fz) = q(E + V x B)
Here E = (0, 0, Eo), V = (Vx, Vy, Vz) and B = (0, -Bo, 0) ==>
V x B = (Vz*Bo, 0, -Vx*Bo)
m*(dV/dt) = q(Vz*Bo, 0, Eo -Vx*Bo)
(m/qBo)*(dV/dt) = (Vz, 0, Eo/Bo -Vx) ==>

(m/qBo)*(dVx/dt) = Vz ..... (1)
(m/qBo)*(dVz/dt) = Eo/Bo -Vx ..... (2)

Differentiate (1) wrt t ==> (m/qBo)*(d²Vx/dt²) = dVz/dt ..... (3)
Substitute dVz/dt from (3) in (2) ==>
(m/qBo)² * (d²Vx/dt²) = Eo/Bo -Vx
(m/qBo)² * (d²Vx/dt²) + Vx = Eo/Bo ..... (4)
The solution of the differential equation (4) is
Vx = (Eo/Bo)*[1-cos(qBo*t/m)] ..... (5)
From (1) ==>
Vz = (Eo/Bo)*sin(qBo*t/m) ..... (6)

From (5) and (6) ==>
X = (Eo/Bo)*[t - (m/qBo)sin(qBo*t/m)] ..... (7)
Z = (mEo/qBo²)*[1 - cos(qBo*t/m)] ..... (8)

From (7) and (8) ==>
The particle is moving along the cycloid in the xz coordinate plane.

V² = Vx² + Vz² = (Eo/Bo)² * [2 - 2cos(qBo*t/m)]
V² = (Eo/Bo)² * 2[1 - cos(qBo*t/m)] ..... (9)
From (8) and (9) ==> V² = 2qEo*Z/m
V = √(2qEo*Z/m)

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