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QPages 230-234: Q2: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q2 Solution: An electron is projected horizontally with a kinetic energy of…

An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic field of strength 1.0×10−7 T1.0 \times 10^{-7} \text{ T} exists in the vertically upward direction. a) Will the electron deflect towards the right or left of its motion? b) Calculate the sideways deflection of the electron while travelling through 1 m. Make appropriate approximations.

Step-by-Step Verified Solution

**Answer:** (a) The electron deflects towards the left. (b) Deflection s≈1.5×10−2 cms \approx 1.5 \times 10^{-2} \text{ cm}. **Step-by-Step Explanation:** (a) The electron moves horizontally and the magnetic field is vertically upward. Using F⃗=−e(v⃗×B⃗)\vec{F} = -e(\vec{v} \times \vec{B}), the force is towards the **left** (if moving in the +x direction). (b) KE=10 keV=1.6×10−15 JKE = 10 \text{ keV} = 1.6 \times 10^{-15} \text{ J}. v=2KE/m=2×1.6×10−15/9.1×10−31=5.93×107 m/sv = \sqrt{2KE/m} = \sqrt{2 \times 1.6 \times 10^{-15} / 9.1 \times 10^{-31}} = 5.93 \times 10^7 \text{ m/s}. The acceleration a=evB/ma = e v B / m. Time to travel 1 m: t=1/vt = 1/v. Deflection s=12at2=eB2mvs = \frac{1}{2} a t^2 = \frac{e B}{2 m v}. Substituting values gives s≈1.5×10−2 cms \approx 1.5 \times 10^{-2} \text{ cm}.

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