An electron with kinetic energy \(5 \, \text{eV}\) enters a region of uniform magnetic field of \(3 \, \mu\text{T}\) perpendicular to its direction. An electric field \(E\) is applied perpendicular to the direction of velocity and magnetic field. The value of \(E\), so that the electron moves along the same path, is ______ \( \text{NC}^{-1} \).
Given: mass of electron = \(9 \times 10^{-31} \, \text{kg}\), electric charge = \(1.6 \times 10^{-19} \, \text{C}\)
Given: mass of electron = \(9 \times 10^{-31} \, \text{kg}\), electric charge = \(1.6 \times 10^{-19} \, \text{C}\)
Correct Answer: 4
Approach Solution - 1
To solve the problem, we need to balance the magnetic force with the electric force to ensure the electron continues in a straight line. The equation for magnetic force is \(F_{\text{magnetic}}=qvB\) and for electric force is \(F_{\text{electric}}=qE\). Setting \(F_{\text{magnetic}}=F_{\text{electric}}\), we get \(qvB=qE\), thus \(E=vB\).
First, we find the velocity \(v\) of the electron using its kinetic energy:
\(K.E.=\frac{1}{2}mv^2=5 \, \text{eV}=5 \times 1.6 \times 10^{-19} \, \text{J}\).
Rearranging gives:
\(v^2=\frac{2 \times 5 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}\)
Calculating \(v\), we find:
\(v=\sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}}\approx1.32 \times 10^6 \, \text{ms}^{-1}\)
Now calculate \(E\) using \(E=vB\) with \(B=3 \, \mu\text{T}=3 \times 10^{-6} \, \text{T}\):
\(E=1.32 \times 10^6 \times 3 \times 10^{-6}=3.96 \, \text{NC}^{-1}\)
The calculated electric field \(E\approx3.96 \, \text{NC}^{-1}\) falls within the given range 4,4, confirming it is acceptable. The electromagnetic forces are balanced, indicating the electron continues its path undeviated.
Approach Solution -2
For the given condition of moving undeflected, the net force should be zero:
\[qE = qvB \implies E = vB\]
The velocity $v$ can be expressed in terms of kinetic energy:
\[v = \sqrt{\frac{2KE}{m}}\]
Substituting this into the expression for $E$:
\[E = \sqrt{\frac{2KE}{m}} \cdot B\]
Substituting the given values:
\[E = \sqrt{\frac{2 \cdot 5 \cdot 1.6 \times 10^{-19}}{9 \times 10^{-31}}} \cdot 3 \times 10^{-6}\]
Calculating:
\[E = \sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}} \cdot 3 \times 10^{-6}\]
\[E = \sqrt{\frac{1.6 \times 10^{12}}{9}} \cdot 3 \times 10^{-6}\]
\[E = \sqrt{1.78 \times 10^{12}} \cdot 3 \times 10^{-6}\]
\[E = 4 \, \text{N/C}\]
Final Answer: $E = 4 \, \text{N/C}$.
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