JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 11)
A metallic rod of length 'L' is rotated with an angular speed of '$$\omega$$' normal to a uniform magnetic field 'B' about an axis passing through one end of rod as shown in figure. The induced emf will be :
_24th_January_Evening_Shift_en_11_1.png)
$$\mathrm{\frac{1}{2}B^2L^2\omega}$$
$$\mathrm{\frac{1}{2}BL^2\omega}$$
$$\mathrm{\frac{1}{4}BL^2\omega}$$
$$\mathrm{\frac{1}{4}B^2L\omega}$$
Explanation
Velocity of centre of $\operatorname{rod} v=\frac{\omega L}{2}$
So, $\mathrm{emf}=B \cdot v L$ $$ =\frac{B \omega L^{2}}{2} $$
So, $\mathrm{emf}=B \cdot v L$ $$ =\frac{B \omega L^{2}}{2} $$
_24th_January_Evening_Shift_en_11_2.png)
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