Practicing Success
Practicing Success
CUET Preparation Today
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$\text{Force applied by electric field of wire on charge }F = \frac{q\lambda}{2\pi \epsilon_0 R}$ $ \text{This force will give centripetal force }\Rightarrow \frac{mv^2}{R} = \frac{q\lambda}{2\pi \epsilon_0 R} $ $\Rightarrow v = \sqrt{\frac{q\lambda}{2\pi \epsilon_0 mR}} = 5.6\times 10^7m/s$ |
