Verified Solution • 2025–26 Session moderate

QPages 230-234: Q48: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q48 Solution: A proton goes undeflected in a crossed electric and magnetic…

A proton goes undeflected in a crossed electric and magnetic field the fields are perpendicular to each other) at a speed of 2.0×105 m s−12.0 \times 10^5 \text{ m s}^{-1}. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and magnetic fields. Take the mass of the proton = 1.6×10−27 kg1.6 \times 10^{-27} \text{ kg}.

Step-by-Step Verified Solution

**Answer:** E=1.0×104 V/mE = 1.0 \times 10^4 \text{ V/m}, B=5.0×10−2 TB = 5.0 \times 10^{-2} \text{ T}. **Step-by-Step Explanation:** For undeflected motion: eE=evB⇒E=vBeE = e v B \Rightarrow E = v B. When electric field is off: r=mveB=0.04 mr = \frac{m v}{e B} = 0.04 \text{ m}. B=mver=1.67×10−27×2×1051.6×10−19×0.04=0.052 TB = \frac{m v}{e r} = \frac{1.67 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.04} = 0.052 \text{ T}. E=vB=2×105×0.052=1.04×104 V/mE = v B = 2 \times 10^5 \times 0.052 = 1.04 \times 10^4 \text{ V/m}.

All Questions in this Chapter

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