ExamPlay Light Logo
登入

JEE MAIN - Physics (2006 - No. 13)

The flux linked with a coil at any instant $$'t'$$ is given by
$$\phi = 10{t^2} - 50t + 250$$
The induced $$emf$$ at $$t=3s$$ is
$$-190$$ $$V$$
$$-10$$ $$V$$
$$10$$ $$V$$
$$190$$ $$V$$

解释

$$\phi = 10{t^2} - 50t + 250$$

$$e = - {{d\phi } \over {dt}} = - \left( {20t - 50} \right)$$

$${e_{t = 3}} = - 10\,V$$

评论 (0)

登录发表评论
广告
BrainBehindX Inc Logo
©2026; 供电 BrainBehindX Inc