Magnetic Field of a Moving Charge Problems and Solutions

 Problem#1

A +6.00 μC point charge is moving at a constant 8.00 x 10⁶ m/s in the +y-direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic-field vector B it produces at the following points: (a) x = 0.500 m, y = 0, z = 0; (b) x = 0, y = ─0.500 m, z = 0;  (c) x = 0, y = 0, z = 5.00 m; (d) x = 0, y = ─5.00 m, z = 5.00 m?

Answer:
Known:
Velocity, v = (8.00 x 10⁶ m/s)j

Magnetic field of a point charge with constant velocity given by

B = (μ₀/4π)(qv x r)/r³

so that, the magnetic-field vector B it produces at the following points:

(a) for x = 0.500 m, y = 0, z = 0; or r = (0.500 m)i is

B = [(4π x 10^─7 T.m/A)/4π]{(6.00 x 10^─6 C)(8.00 x 10⁶ m/s)j x (0.500 m)i/(0.500 m)³
B = (─1.92 x 10^─5 T)k

(b) for x = 0, y = ─0.500 m, z = 0; or r = (─0.500 m)j is

B = [(4π x 10^─7 T.m/A)/4π]{(6.00 x 10^─6 C)(8.00 x 10⁶ m/s)j x (─0.500 m)j/(0.500 m)³
B = 0

(c) for x = 0, y = 0, z = 0.500 m; or r = (0.500 m)k is

B = [(4π x 10^─7 T.m/A)/4π]{(6.00 x 10^─6 C)(8.00 x 10⁶ m/s)j x (0.500 m)k/(0.500 m)³
B = (1.92 x 10^─5 T)i

(d) for x = 0, y = ─5.00 m, z = 5.00 m; or r = (─0.500 m)j + (0.500 m)k and r = 0.500√2 m   is

B = [(4π x 10^─7 T.m/A)/4π]{(6.00 x 10^─6 C)(8.00 x 10⁶ m/s)j x (0.500 m)(k ─ j)/(0.500√2 m)³
B = (6.79 x 10^─6 T)i

Problem#2
Fields Within the Atom. In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.3 x 10^-11 m with a speed of 2.2 x 10⁶ m/s. If we are viewing the atom in such a way that the electron’s orbit is in the plane of the paper with the electron moving clockwise, find the magnitude and direction of the electric and magnetic fields that the electron produces at the location of the nucleus (treated as a point).

Answer:
Magnetic field of a point charge with constant velocity given by
B = (μ₀/4π)(qv x r)/r³
B = (μ₀/4π)(qv sin φ)/r²

Therefore:
B = [(4π x 10^─7 T.m/A)/4π]{(1.60 x 10^─19 C)(2.20 x 10⁶ m/s) x sin 90⁰/(5.3 x 10^-11 m)²
B = 13.0 T (out of the page)

and its electric field given by
E = ke/r²
   = (9.0 x 10⁹ Nm²/C²)(1.60 x 10^─19 C)/(5.3 x 10^-11 m)²
E = 5.1 x 10 ¹¹ N/C (toward the electron)

Problem#3
An electron moves at 0.100c as shown in Fig. 1. Find the magnitude and direction of the magnetic field this electon produces at the following points, each 2.00 μm from the electron: (a) points A and B; (b) point C; (c) point D.

Fig.1

Answer:
Magnetic field of a point charge with constant velocity given by
B = (μ₀/4π)(qv x r)/r³
B = (μ₀/4π)(qv sin φ)/r²

(a) at points A, the magnitude and direction of the magnetic field this electon is
B = [(4π x 10^─7 T.m/A)/4π]{(1.60 x 10^─19 C)(0.100 x 3.0 x 10⁸ m/s) x sin 30⁰/(2.00 x 10^-6 m)²
B = 6.00 x 10^─8 T, out of the paper and it’s the same at point B.

(b) at points C, the magnitude and direction of the magnetic field this electon is
B = [(4π x 10^─7 T.m/A)/4π]{(1.60 x 10^─19 C)(0.100 x 3.0 x 10⁸ m/s) x sin 90⁰/(2.00 x 10^-6 m)²
B = 1.20 x 10^─7 T, out of the paper

(c) at points D, the magnitude and direction of the magnetic field this electon is
B = [(4π x 10^─7 T.m/A)/4π]{(1.60 x 10^─19 C)(0.100 x 3.0 x 10⁸ m/s) x sin 180⁰/(2.00 x 10^-6 m)²
B = 0

Problem#4
An alpha particle (charge ) and an electron move in opposite directions from the same point, each with the speed of 2.50 x 10⁵ m/s (Fig. 2). Find the magnitude and direction of the total magnetic field these charges produce at point P, which is 1.75 nm from each of them.

Fig.2

Answer:
Magnetic field of a point charge with constant velocity given by
B = (μ₀/4π)(qv sin φ)/r²

Both moving charges produce magnetic fields, and the net field is the vector sum of the two fields. Therefore,

B_net = B_alpha + B_el
With, B_alpha = (μ₀/4π)(2ev sin 140⁰)/r² (out the paper)
and B_el = (μ₀/4π)(ev sin 40⁰)/r² (out the paper), then
B_net = (μ₀/4π)(ev/r²)(2sin 140⁰ + sin 40⁰)
B = [(4π x 10^─7 T.m/A)/4π][(1.60 x 10^─19 C)(2.50 x 10⁵ m/s)/(1.75 x 10^-9 m)²](2sin 140⁰ + sin 40⁰)
B = 2.52 x 10^─3 T, out of the paper

Problem#5
A ─4.80 μC point charge is moving at a constant 6.80 x 10⁵ m/s in the +x-direction, relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic-field vector it produces at the following points: (a) x = 0.500 m, y = 0, z = 0; (b) x = 0, y = 0.500 m, z = 0;  (c) x = 5.00 m, y = 5.00 m, z = 0; (d) x = 0, y = 0, z = 5.00 m?

Answer:
Known:
Velocity, v = (6.80 x 10⁵ m/s)i

Magnetic field of a point charge with constant velocity given by
B = (μ₀/4π)(qv x r)/r³

so that, the magnetic-field vector B it produces at the following points:

(a) for x = 0.500 m, y = 0, z = 0; or r = (0.500 m)i is
B = [(4π x 10^─7 T.m/A)/4π]{(4.80 x 10^─6 C)(6.80 x 10⁵ m/s)i x (0.500 m)i/(0.500 m)³
B = 0

(b) for x = 0, y = 0.500 m, z = 0; or r = (0.500 m)j is
B = [(4π x 10^─7 T.m/A)/4π]{(─4.80 x 10^─6 C)(6.80 x 10⁵ m/s)i x (0.500 m)j/(0.500 m)³
B = (─1.31 x 10^─6 T)i

(c) for x = 5.00 m, y = 5.00 m, z = 0; or r = (0.500 m)i + (0.500 m)j and r = 0.500√2 m is
B = [(4π x 10^─7 T.m/A)/4π]{(─4.80 x 10^─6 C)(6.80 x 10⁵ m/s)i x (0.500 m)(i + j)/(0.500 m√2)³
B = (─4.62 x 10^─7 T)k

(d) for x = 0, y = 0, z = 5.00 m; or r = (0.500 m)k and r = 0.500 m   is
B = [(4π x 10^─7 T.m/A)/4π]{(─4.80 x 10^─6 C)(6.80 x 10⁵ m/s)i x (0.500 m)k/(0.500 m)³
B = (1.31 x 10^─6 T)j  

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