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Question: An electron accelerated by a potential difference V=1.0 kV moves in a uniform magnetic field at an a...

An electron accelerated by a potential difference V=1.0 kV moves in a uniform magnetic field at an angle α=30∘\alpha = {30^\circ } to the vector B whose modulus is B=29  mTB = 29\;{\rm{mT}}. Find the pitch of the helical trajectory of the electron.
A. 1 m
B. 1 cm
C. 2 m
D. 2 cm

Explanation

Solution

when an electron is placed in a magnetic field then it will experience some field and moves with some velocity in a particular path, so such motion in a particular path is called the trajectory of the electron.

Complete step by step answer:
When an electron is placed in a magnetic field then it will experience some field and moves with some velocity in a particular path, so such motion in a particular path is called the trajectory of the electron.
Complete step by step answer
Given:
The potential difference of the electron is V=1.0  kVV = 1.0\;{\rm{kV}}.
The angle of the electron while moving in a uniform magnetic field is α=30∘\alpha = 30^\circ .
The vector modulus is B=29  mTB = 29\;{\rm{mT}}
The equation to find the trajectory of the electron is,.
P=vHT=2πmBe2eVmcos⁡αP = {v_H}T = \dfrac{{2\pi m}}{{Be}}\sqrt {\dfrac{{2eV}}{m}} \cos \alpha
Here, p is the pitch of the helical trajectory; v is the velocity in horizontal direction.
The standard value of the e electron charge is1.6×10−19  C1.6 \times {10^{ - 19}}\;{\rm{C}}
The standard value of the m is 9.1×10−31  kg9.1 \times {10^{ - 31}}\;{\rm{kg}}
Substitute the values in the above equation.
P=vHT =2πmBe2eVmcos⁡α =2×3.14×(9.1×10−31  Kg) 29  mT×1.6×10−19 C2×(1.6×10−19 C)×(103  V)9.1×10−31  Kgcos⁡30∘ =0.02  m ×100  cm1  m =2cm\begin{array}{l} P = {v_H}T\\\ = \dfrac{{2\pi m}}{{Be}}\sqrt {\dfrac{{2eV}}{m}} \cos \alpha \\\ = \dfrac{{2 \times 3.14 \times \left( {9.1 \times {{10}^{ - 31}}\;{\rm{Kg}}} \right)\,}}{{29\;{\rm{mT}} \times 1.6 \times {{10}^{ - 19}}\,{\rm{C}}}}\sqrt {\dfrac{{2 \times \left( {1.6 \times {{10}^{ - 19}}\,{\rm{C}}} \right) \times \left( {{{10}^3}\;{\rm{V}}} \right)}}{{9.1 \times {{10}^{ - 31}}\;{\rm{Kg}}}}} \cos 30^\circ \\\ = 0.02\;{\rm{m}}\, \times \dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}\\\ = {\rm{2 cm}} \end{array}

So, the correct answer is “Option D”.

Note:
Here, the standard values should be remembered and maintain the accuracy at the values. Make sure that all the values are in equal standard so the output will not be affected by that.