Problem#1
A flat, square surface with side length 3.40 cm is in the xy-plane at . Calculate the magnitude of the flux through this surface produced by a magnetic field B = (0.200 T)i + (0.300 T)j – (0.500T)kAnswer:
Known:
Area, A = (0.024 m)2k
magnetic field, B = (0.200 T)i + (0.300 T)j – (0.500T)k
the magnitude of the flux through this surface produced by a magnetic field is
φ = B.A
φ = ((0.200 T)i + (0.300 T)j – (0.500T)k)[(0.024 m)2k]
φ = –0.012 Wb
Problem#2
A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field (a) in the +z-direction; (b) at an angle of from the +z-direction; (c) in the direction?
Answer:
Known:
Area, A = π x (0.065 m)2 k = 0.0132k
Magnetic Flux, φ = B.A
(a) uniform magnetic field in the +z-direction (B = 0.23k)
φ = B.A = (0.23k)(0.0132k) = 3.05 x 10-3 Tm2
(b) uniform magnetic field at an angle of from the +z-direction (B = 0.23 x cos 530 k)
φ = B.A = (0.23k)(0.23 x cos 530 k) = 1.83 x 10-3 Tm2
(c) uniform magnetic field in the direction (B = 0.23j)
φ = B.A = (0.23k)(0.23j) = 0
Probme#3
A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle 30.00 of above the horizontal. What must the magnitude of the magnetic field be in order to produce a flux of 4.20 x 10-4 Wb through the surface?
Answer:
Known:
Area, A = 0.0280 m x 0.0320 m
Flux, φ = 4.20 x 10-4 Wb
where, θ = 60.00
Solving we can get
φ = BA cos θ
for B gives,
B = φ/(A cos θ)
B = (4.20 x 10-4 Wb)/[0.0280 m x 0.0320 m x cos 60.00]
B = 0.938 T
Problem#4
An open plastic soda bottle with an opening diameter of 2.5 cm is placed on a table. A uniform 1.75-T magnetic field directed upward and oriented 250 from vertical encompasses the bottle. What is the total magnetic flux through the plastic of the soda bottle?
Answer:
Known:
magnetic field, B = 1.75 T
Area, A = ¼ πd2 = ¼ π x (0.0250 m)2
Angle, θ = 250
the total magnetic flux through the plastic of the soda bottle is
φ = BA cos θ
φ = (1.75 T)(¼ π x (0.0250 m)2)(cos 25.00) = 7.785 x 10-4 Wb
Problem#5
The magnetic field in a certain region is 0.128 T, and its direction is that of the +z-axis in Fig. 1. (a) What is the magnetic flux across the surface abcd in the figure? (b) What is the magnetic flux across the surface befc? (c) What is the magnetic flux across the surface aefd? (d) What is the net flux through all five surfaces that enclose the shaded volume?
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| Fig.1 |
Answer:
Known:
magnetic field, B = 0.128 T direction is that of the +z-axis (B = 0.128k)
(a) is the magnetic flux across the surface abcd is
φ = B.A
where, Aabcd = 0.4 m x 0.3 m k = 0.12 m2i
φ = (0.128k).(0.12 m2i) = 0
(b) is the magnetic flux across the surface befc is
φ = B.A
where, Abefc = 0.3 m x 0.3 m k = 0.09 m2k
φ = (0.128k).(0.02 m2k) = 2,5 mWb
(b) is the magnetic flux across the surface aefd is
φ = B.A
where, Aaefd = 0.3 m x 0.5 m = 0.15 m2
φ = 0.128 k x 0.15 m2 cos 530 k = 0.0113 Wb
magnetic field, B = 0.128 T direction is that of the +z-axis (B = 0.128k)
(a) is the magnetic flux across the surface abcd is
φ = B.A
where, Aabcd = 0.4 m x 0.3 m k = 0.12 m2i
φ = (0.128k).(0.12 m2i) = 0
(b) is the magnetic flux across the surface befc is
φ = B.A
where, Abefc = 0.3 m x 0.3 m k = 0.09 m2k
φ = (0.128k).(0.02 m2k) = 2,5 mWb
(b) is the magnetic flux across the surface aefd is
φ = B.A
where, Aaefd = 0.3 m x 0.5 m = 0.15 m2
φ = 0.128 k x 0.15 m2 cos 530 k = 0.0113 Wb
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