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Step-by-Step Detailed Solution
Step 1: Identify the Known Parameters
• Radius of circular path, $r = 0.5\,\text{cm} = 0.5 \times 10^{-2}\,\text{m}$
• Magnetic field, $B = 0.5\,\text{T}$
• Electric field, $E = 100\,\text{V/m}$
• Charge of the particle (same as electron), $q = 1.6 \times 10^{-19}\,\text{C}$
Step 2: Write Down the Relevant Equations of Motion
1. When a charged particle moves in a circle under a uniform magnetic field, the magnetic force provides the centripetal force:
$ \displaystyle q v B = \frac{m v^2}{r} \quad\Longrightarrow\quad r = \frac{m v}{q B} \quad (1)$
2. When both electric and magnetic fields are applied in such a way that the particle moves in a straight line, the net force on the particle is zero. Thus, the electric force must balance the magnetic force:
$ \displaystyle qE = q v B \quad\Longrightarrow\quad E = v B \quad (2)$
Step 3: Relate the Velocity from the Electric and Magnetic Fields
From equation (2), solve for $v$:
$ \displaystyle v = \frac{E}{B}.$
Step 4: Substitute Velocity into the Expression for Radius
Using equation (1) and substituting $v = \tfrac{E}{B}$, we get:
$ \displaystyle r = \frac{m\left(\tfrac{E}{B}\right)}{q B} = \frac{m E}{q B^2}.$
Hence, solving for $m$:
$ \displaystyle m = \frac{q B^2\,r}{E}.$
Step 5: Substitute the Numerical Values
$ \displaystyle m = \frac{\bigl(1.6 \times 10^{-19}\,\text{C}\bigr)\,\bigl(0.5\,\text{T}\bigr)^2 \,\bigl(0.5 \times 10^{-2}\,\text{m}\bigr)}{100\,\text{V/m}}. $
Calculate step by step:
$ B^2 = (0.5)^2 = 0.25\,\text{T}^2. $
$ q B^2 = 1.6 \times 10^{-19} \times 0.25 = 0.4 \times 10^{-19}. $
$ q B^2\,r = 0.4 \times 10^{-19} \times 0.5 \times 10^{-2} = 0.4 \times 0.5 \times 10^{-21} = 0.2 \times 10^{-21} = 2.0 \times 10^{-22}. $
$ m = \frac{2.0 \times 10^{-22}}{100} = 2.0 \times 10^{-24}\,\text{kg}. $
Step 6: State the Final Answer
Therefore, the mass of the particle is $2.0 \times 10^{-24}\,\text{kg}$.
