Question

A charge particle of $2 \,\mu C$ accelerated by a potential difference of $100\, V$ enters a region of uniform magnetic field of magnitude $4 \,mT$ at right angle to the direction of field The charge particle completes semicircle of radius $3 \,cm$ inside magnetic field The mass of the charge particle is _______ $\times 10^{-18} kg$

Answer 1

Correct Answer: 144

To find the mass of the charged particle, we'll use principles from electromagnetism. Given: Charge (q) = 2 μC = 2 × 10^-6 C, Potential difference (V) = 100 V, Magnetic field (B) = 4 mT = 4 × 10^-3 T, Radius (r) = 3 cm = 0.03 m.
First, calculate the velocity (v) of the particle using the energy equation: qV = 0.5mv². Rearranging gives: v = √(2qV/m). However, we need the mass.
Next, use the formula for the radius of circular motion in a magnetic field: r = mv/(qB). Rearrange to solve for mass (m): m = rqB/v.
Since v = √(2qV/m), substitute into m:
Let u = √(2qV/m) so, m = rqB/u.
Thus, u = √(2qV/(rqB)).
Substitute numerical values:
Calculate kinetic energy: K.E. = qV = (2 × 10^-6 C)(100 V) = 2 × 10^-4 J.
The velocity from energy: v = √((2 × 2 × 10^-4 J)/m).
Substitute v into the radius expression: m = rqB × (m)/(√(2r²qB)).
Simplify: m = 2rqB × m/qB.
Cancel: m = 2rqB × 1/qB = 2r.
Put the values: m = 2 × 0.03 × (4 × 10^-3 T)/(2 × 10^-6 C).
Calculated mass: m = 1.44 × 10^-16 kg.
Express in terms of 10^-18: m = 144 × 10^-18 kg.
Check the range: min = 144, max = 144. Thus, m fits the range.