A proton moving with a velocity 3 into 10 to the power 5 m per second enters in a magnetic field of 0.3 tesla at an angle 30 degree with the field the radius of curvature of its path will be
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4
velocity of proton have two components
vcostheta and vsintheta
as we know
m(vsintheta)^2/r=qvBsintheta
r=mvsintheta/qB
r=1.67×10^-27×3×10^5×sin30/1.6×10^-19×0.3
r=3.81/0.96×10^-3
r=3.9×10^-3
vcostheta and vsintheta
as we know
m(vsintheta)^2/r=qvBsintheta
r=mvsintheta/qB
r=1.67×10^-27×3×10^5×sin30/1.6×10^-19×0.3
r=3.81/0.96×10^-3
r=3.9×10^-3
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14
Answer:
F = q v B sinθ = mv²/ R
where, q is the charge, v is the velocity, B is the strength of magnetic field, m is the mass and R is the radius of curvature.
R = (mv)/(qB sinθ)
m = 1.67 ×10⁻²⁷ kg
v = 3 × 10⁵ m/s
B = 0.3 T
θ = 30°
q = 1.6 ×10⁻¹⁹ C
⇒ R = (1.67 ×10⁻²⁷ kg×3 × 10⁵ m/s)/(1.6 ×10⁻¹⁹ C× 0.3 T × sin 30°)
⇒R = 20.875 × 10 ⁻³ m = 20.875 mm.
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