A proton moving towards east in a horizontal plane enters a horizontal magnetic field of 0.34 t directed towards north with a speed of 2×10^7 m/s .Calculate the lateral displacement of the proton while moving 0.20m towards east.
Answers
Answer:
Explanation:
Answer:
Explanation:
=> Let the north direction be jˆ and the east direction be iˆ
v=2.0×10⁷ iˆ m/s
B=0.34jˆT
q=e=1.602×10⁻¹⁹C
F=q(v×B)
m_proton=1.674×10⁻²⁷kg
F=1.602×10⁻¹⁹(2.0×10⁷ iˆ×0.34jˆ)
=1.09×10⁻¹² kˆ (Note: kˆ points vertically upwards )
R=mv/qB
=1.674×10⁻²⁷×2×10⁷/1.602×10⁻¹⁹×0.34
=0.625m
=> the lateral displacement of the proton while moving 0.20m towards east can be calculated if we calculate the distance of proton from horizontal axis when its distance from vertical axis is 0.2m.
As shown in figure, ΔCPQ is a right angle triangle
According to Pythagoras theorem,
CP² = CQ² + QP²
CQ² = CP² - QP²
CQ² = R² - (0.2)²
CQ² = (0.625)² - (0.2)²
CQ² = 0.351
CQ = √0.351
= 0.592m
Lateral displacement = Distance of particle from horizontal axis = R - CQ = 0.625 - 0.592 = 0.032m
Thus, the lateral displacement of the proton while moving 0.20m towards east is 0.032m.