An alpha-particle moves with a speed of $(5×10^5\hat i) m s^{-1}$. It enters a region where there is a magnetic field of magnitude 4 T, directed at an angle of 45° to the X-axis and lying in XY plane. The magnitude of magnetic force on alpha particle is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $4.5 × 10^{-13} N$
To calculate the Magnetic field ($\vec B$) on alpha-particle,
$F=q(\vec v×\vec B)$
$=qvB\sin θ$
where,
q = charge on particle = $2e = 2×1.6×10^{-19}C$
v = velocity of α-particle = $5×10^5m/s$
B = magnetic field = 4T
$θ=45°$ (angle between $\vec v$ and $\vec B$)
$∴F=(2×1.6×10^{-19})×(5×10^5)×(4)×(\frac{1}{\sqrt{2}})$
$=\frac{64}{\sqrt{2}}×10^{-14}$
$≃4.5 × 10^{-13} N$