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Question 84

(a) A circular conducting loop carries a current of 5.0 A and has radius 30 mm
(i) Calculate the magnetic field at the centre of the loop
(H) How many turns would be needed to give a field of 1.0 mT
(b) A electron is injected into the 1.0 mT magnetic field mentioned in part (a) with a velocity of 2500 m S-lin a direction perpendicular to the field.
(i)What is the force on the electron?
(H)Describe the nature and magnitude of the path of the electron due to the influence of the magnetic field

The diagram above shows the end of a fiber optic cable with core refractive index 1.486 and cladding refractive index 1.472

(a) calculate the critical angle at which light in the core is just totally internally reflected at X
(b) calculate the angle of incidence at A for light which is just internally reflected in the cable
(c) calculate the speed of light in the core
(d) calculate the time taken for light to travel 1.0 km directly down the center of the core
(e) calculate the longest time taken for light to traverse 1.0 km of cable by internal reflection
(t) estimate the maximum frequency of a signal which can be transmitted down 1.0 km of this optical fiber

Question 86

(a) Kr-85 is a radioactive isotope that is used in measuring paper thickness. It decays by If emission with a half-life of 10.8 years
(i) write down the decay equation showing the daughter isotope and decay products of this process
(H)a Kr-85 source had activity 30 mCi when new. Estimate its activity after 5 years' use
(iii) what is the activity of a 30 mCi source in becquerels?
(iv) what is the mass of Kr-85 in the source when new?

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B4)
a)
current i = 5.0 A
radius r = 30 mm = 0.030 m

i.)
Magnetic field at the center:
B = k*2*pi*i/r
where, l = mu_o/4pi = 10^(-7) N/A^2
Hence,
B = 10^(-7)*2*pi*5/0.03 = 1.05*10^(-4) T = 1.05 Gauss --Answer

ii)
If number of turns = N
Hence,
B = N*1.05*10^(-4) = 1mT = 1*10^(-3) T
=> N = 10.5 == 11 turns --Answer

b)
B = 1mT = 1*10^(-3) T
v = 2500 m/s
Because,
F = q*v*B
where, charge, q = e = 1.6*10^(-19) C

Hence,
F = 1.6*10^(-19)* 2500 *1*10^(-3) = 4*10^(-19) N --Answer
The force will ...

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  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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