Answer:
Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T
Explanation:
Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :
......(1)
Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.
In this problem,
Current, I = 0.7 A
Length of wire, L = 0.62 m
Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m
Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m
Substitute these values in equation (1).
B = 6.99 x 10⁻⁶ T
Answer:
a). V = 3.13*10⁶ m/s
b). T = 1.19*10^-7s
c). K.E = 2.04*10⁵
d). V = 1.02*10⁵V
Explanation:
q = +2e
M = 4.0u
r = 5.94cm = 0.0594m
B = 1.10T
1u = 1.67 * 10^-27kg
M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg
a). Centripetal force = magnetic force
Mv / r = qB
V = qBr / m
V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27
V = 2.09088 * 10^-20 / 6.68 * 10^-27
V = 3.13*10⁶ m/s
b). Period of revolution.
T = 2Πr / v
T = (2*π*0.0594) / 3.13*10⁶
T = 1.19*10⁻⁷s
c). kinetic energy = ½mv²
K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²
K.E = 3.27*10^-14J
1ev = 1.60*10^-19J
xeV = 3.27*10^-14J
X = 2.04*10⁵eV
K.E = 2.04*10⁵eV
d). K.E = qV
V = K / q
V = 2.04*10⁵ / (2eV).....2e-
V = 1.02*10⁵V
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