Question 1 of 1 A charged particle is moving in a circular path with velocity v in a uniform magnetic field B, if the velocity of charged particle is doubled and strength of magnetic field is halved, then radius becomes A 8 times B 4 times C 2 times D 16 times 💡 Explanation Radius of circular motion: [latex]r = \dfrac{mv}{qB}[/latex]. Doubling v and halving B: [latex]r' = \dfrac{m(2v)}{q(B/2)} = 4 \times \dfrac{mv}{qB} = 4r[/latex]. So the radius becomes 4 times — option (b). ← Prev Check Next → Finish & See Score 0 of 1 checked ← Back 1 / 1 1
