Question

QUESTION 9

1. A 1,000-turn coil has a cross section of 7.0 cm ^ 2 and a length of 25 cm. Determine how much energy is stored in the coil's magnetic field when it has a current of 10.0 A.

a. 0.18 J

b. 0.36 J

c. 0.10 J

d. 28 J

e. 2.8 J

QUESTION 10

1. Determine which of the following types of waves is intrinsically different from the other four.

a. ultraviolet radiation

b. gamma rays

c. radio waves

d. visible light

e. sound waves

QUESTION 11

1. A circular copper cable is located perpendicular to a uniform magnetic field of 0.50 T. Due to external forces the cable area decreases at a ratio of 1.26 x 10 ^ -3 m ^ 2 / s. Determine the "emf" induced in the circular cable.

a. 7.9 x 10 ^ -3 V

b. 3.1 V

c. 1.2 x 10 ^ -3 V

d. 6.3 x 10 ^ -4 V

e. 3.1 x 10 ^ -4 V

QUESTION 12

1. A cable consists of 240 circular turns, each radius 0.044 m, and the current is 2.2 A. Determine the magnetic moment of the cable.

a. 23 Am ^ 2

b. 0.65 Am ^ 2

c. 15 am ^ 2

d. 0.21 Am ^ 2

e. 3.2 Am ^ 2

Answer 1

Q-9
Given
coil with N= 1000 turns with cross section Area of A = 7.0 cm^2 = 7.0*10^-4 m^2 about lenght L = 25 cm = 0.25 m
current in the coil is I = 10.0 A
Energy stored in the coils in magnetic field is E = ?

we know that the energy stores is E = u*V
u is energy density , u = B^2/2*mue0

B is magnetic field , B = mue0*n*I; n = N/L

   B = 4pi*10^-7(1000/0.25)(10.0) T

   B = 0.0503 T

now the energy density is u = B^2/(2*mue0)

total energy is E = u*V = B^2/(2*mue0)*V = B^2/(2*mue0)(A*L)

E = ((0.0503^2)/(2*4pi*10^-7))(7*10^-4)(0.25) J

E = 0.18 J
Answer is option (a) 0.18 J

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Q-10
Sound waves are different from the given waves

Sound waves are longitudinal waves where as others were transverse waves, sound waves will not be a part of electromagnetic spectrum.

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Q-11
Given Magnetic field is B = 0.50 T
rate of area decreasing is -dA/dt= 1.26 x 10 ^ -3 m ^ 2 / s.

the "emf" induced in the circular cable is e = ?

we know that the induced emf, e = -d(phi/dt)
where phi is magnetic flux , phi = B*A

   e = - B*(-dA/dt) = 0.5*1.26*10^-3 V

   e = 0.00063 V

   e = 6.3*10^-4 V
Answer is option (d)
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Q-12

cable with N = 240 circular turns
with radius r = 0.044 m
current is 2.2 A.

the magnetic moment of the cable mue = ?

we know that mue = N*I*A

   mue = 240*2.2*pi*0.044^2 A.m^2
   mue = 3.2 A.m^2
Answer is option (e)