Chapter 21
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1). Force acting to the moving charge is given by equation 21.1
F = qvB sin(theta) =>
sin(theta) = F/q*v*B =>
sin(theta) = 2.3*10^(-7)N /(1.7*10^(-5)C * 280 m/s * 5*10^(-5)T) =
= 0.966 =>
angle theta is equal either to 75.1 or to 104.9 degrees.
2). Using Right Hand Rule #1 we can determine that the force acting
on the charge will always be directed into the page.
Magnitude can be computed from equation
F = q*v*B*sin(theta)
a). F = 8.4 microC * 45 m/s * 0.3 * sin30 = 5.4 *10^(-5)N
b). F = 8.4 microC * 45 m/s * 0.3 * sin90 = 1.08*10^(-4)N
c). F = 8.4 microC * 45 m/s * 0.3 * sin150 = 5.4 *10^(-5)N
11). The radius of the circular motion in the magnetic field is given by
the equation 21.2
r = mv/qB,
where r is the radius of the circle, m is the mass of the electron,
v is the electron velocity, q is the electron charge and B is
the magnitude of the magnetic field. Inverting this equation we get
b). B = mv/qr
[9.1*10^(-31)kg*6*10^(6)m/s]/[1.6*10(-19)C*1.3*10^(-3)m] =
= 2.63 * 10^(-2) T
c). The acceleration for the circular motion is given by
a = v^2/r = [6*10^(6) m/s]^2/[1.3*10^(-3)m] = 2.8*10^(16)m/s^2
31). The magnetic force acting on the wire segment carrying a current is
F= I*L*B*sin(theta)
Thus the force acting on the sides of the square that
are parallel to the magnetic field is equal to zero since
the sinus of a zero degree angle between the current and
the magnetic field (parallel) is zero. For the sides
that are perpendicular to the field the corresponding angle is
90 degrees and the sinus of that angle is equal to 1.
The magnitute of the force is therefore
F = I*L*B = 12A * 0.32m * 0.25T = 0.96 N