Question
A charged particle moving in a uniform magnetic field and losses $4 \%$ of its $\mathrm{KE}$. The radius of curvature of its path changes by:(a) $2 \%$(b) $4 \%$(c) $10 \%$(d) none of these
Step 1
We know that the kinetic energy (KE) is given by the formula $\frac{1}{2}mv^2$. The percentage change in kinetic energy can be written as: \[\Delta KE = 2 \times \frac{\Delta v}{v}\] Show more…
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A charged particle moving in a uniform magnetic field when losses $4 \%$ of its kinetic energy, the radius of curvature of its circular path (a) decreases by $2 \%$ (b) increases by $2 \%$ (c) increases by $4 \%$ (d) decreases by $4 \%$
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