1
JEE Main 2023 (Online) 24th January Evening Shift
+4
-1 The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by

$$\mathrm{{E_x} = {E_o}\sin (kz - \omega t)}$$

$$\mathrm{{B_y} = {B_o}\sin (kz - \omega t)}$$

Then the correct relation between E$$_0$$ and B$$_0$$ is given by

A
$$\mathrm{{E_o}{B_o} = \omega k}$$
B
$$\mathrm{{E_0} = k{B_0}}$$
C
$$\mathrm{k{E_0} = \omega {B_0}}$$
D
$$\mathrm{\omega {E_0} = k{B_0}}$$
2
JEE Main 2023 (Online) 24th January Evening Shift
+4
-1 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 : A
$$\mathrm{\frac{1}{2}B^2L^2\omega}$$
B
$$\mathrm{\frac{1}{2}BL^2\omega}$$
C
$$\mathrm{\frac{1}{4}BL^2\omega}$$
D
$$\mathrm{\frac{1}{4}B^2L\omega}$$
3
JEE Main 2023 (Online) 24th January Morning Shift
+4
-1 A conducting circular loop of radius $$\frac{10}{\sqrt\pi}$$ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :

A
emf = 10 mV
B
emf = 5 mV
C
emf = 100 mV
D
emf = 1 mV
4
JEE Main 2023 (Online) 24th January Morning Shift
+4
-1 In $$\overrightarrow E$$ and $$\overrightarrow K$$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by :

($$\omega$$ - angular frequency) :

A
$${1 \over \omega }\left( {\overline K \times \overline E } \right)$$
B
$$\overline K \times \overline E$$
C
$$\omega \left( {\overline K \times \overline E } \right)$$
D
$$\omega \left( {\overline E \times \overline K } \right)$$
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