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