CBSE 2025, Set 1

Question

A charged particle gains a speed of $10^6$ ms$^{-1}$, when accelerated from rest through a potential difference 10 kV. It enters a region of magnetic field of 0·4 T such that $\vec{v} \perp \vec{B}$. The radius of circular path described by it is : A). $2\cdot5$ cm B). $5$ cm C). $8$ cm D). $10$ cm

Edvaya · Accepted Answer

The correct answer is $5$ cm. **Step 1: Find the charge-to-mass ratio.** From the work-energy theorem: $$qV = \frac{1}{2}mv^2 \implies \frac{q}{m} = \frac{v^2}{2V} = \frac{(10^6)^2}{2 	imes 10^4} = \frac{10^{12}}{2 	imes 10^4} = 5 	imes 10^7 	ext{ C/kg}$$ **Step 2: Find the radius of the circular path.** For circular motion in a magnetic field: $$r = \frac{mv}{qB} = \frac{v}{(q/m) \cdot B} = \frac{10^6}{5 	imes 10^7 	imes 0 \cdot 4}$$ $$r = \frac{10^6}{2 	imes 10^7} = 0 \cdot 05 	ext{ m} = \boxed{5 	ext{ cm}}$$ > **Key formulae:** $r = mv/qB = \sqrt{2mV/q}/B$. The ratio $q/m = 5 	imes 10^7$ C/kg is close to that of a proton ($\approx 9.6 	imes 10^7$ C/kg), so this is likely a proton-like particle.

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Physics > Magnetism Induction And Alternating Current > Moving Charges And Magnetism

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