(a) Sketch and label the vacuum tube apparatus which J. J. Thomson used to inves

Question

(a) Sketch and label the vacuum tube apparatus which J. J. Thomson used to investigate the balancing of forces exerted on a charge deriving from electric and magnetic fields. [2 marks] (b) With the aid of equations for forces on a charged particle due to an electric and a magnetic field, determine how the Thomson apparatus may be used to finely select a velocity for a charged particle. Define all variables used and show vectors with a sketch. [2 marks] (c) An electron is accelerated from absolute rest by a potential difference of 300 V. It ther enters a uniform magnetic field of magnitude 100 mT with its velocity perpendicular to the field. By equating the magnetic force on a charged particle with the centripetal force, determine the radius of its path in the magnetic field. [2 marks] (d) Two wires are 3.0 m long and 8.0 mm apart. Each carries a current of 5.0 A in the same on the other. Define all variables used and show vectors with a sketch.[2 marks] direction. Determine the magnitude and direction of the net force that one wire exerts (e) A coil has 20 loops and is rectangular in shape with sides of 5 cm and 10 cm. Each loop carries a current of 3.0 A, and the coil is placed in a magnetic field of 2.0 T. Starting from F= il×B, determine the maximum and minimum magnitudes of torque that could be exerted on the coil. Define all variables used and show vectors with a sketch. [2 marks]

Explanation / Answer

a) You can draw this sketch easily as this is a standard apparatus and can be found in the textbooks easily

b) Now, Here we have two forces, Magnetic forve (FB) and electric force (FB)

Where FB = qvB , FE = qE

so, if we equate these two forces,

qvB = qE

v = E/B

also, qV = 1/2*m*v2 where v is velocity and V is potential difference, m is mass of charged particle

c) first we need to find velocity,

v = sqrt (2qV / m)

v = 1.027e7 m/s

Now, qvB = mv2 / r

qB = mv/r

r = mv/qB

r = 9.1e-31*1.027e7 / 1.6e-19*100e-3

r = 5.8416e-4 m

d) F = uoI1I2L / 2*pi*r

F = 4*pi*10-7*5*5*3 / 2*pi*8e-3

F = 1.875e-3 N

The wires will attract each other

If you need force per unit length, then use the following formula

F/L = uoI1I2 / 2*pi*r

Note that this formula is usually used for very long wires

e)

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