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A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×10⁵ m s⁻¹. When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of the electric field is x×10⁴ N/C. The value of x is ______. Take the mass of the proton =1.6×10⁻²⁷ kg.

Asked in JEE Main 22nd Jan 2nd Shift 2025 · Crossed electric and magnetic fields

Answer: 2

Step-by-step solution

Circular path with E off: B=(mv)/(qr)=(1.6×10⁻²⁷×2×10⁵)/(1.6×10⁻¹⁹×0.02)=0.1 T

Undeflected motion: qE=qvB⇒ E=vB

E=2×10⁵×0.1=2×10⁴ N/C

→ 2

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