1
TS EAMCET 2022 (Online) 18th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

A planet has magnetic dipole moment of $27 \times 10^{22} \mathrm{~A}-\mathrm{m}^2$. If the radius of the planet is 300 km , what would be the magnetic field at its equator? $\left(\right.$ use,$\left.\frac{\mu}{4 \pi}=10^{-7}\right)$

A

1 T

B

27 T

C

11 T

D

30 T

2
TS EAMCET 2022 (Online) 18th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

A long solenoid has 20 turns per cm. A small loop of area $4 / \pi \mathrm{cm}^2$ is placed inside the solenoid normal to its axis. If the current carried by the solenoid changed steadily from 1.0 A to 3.0 A in 0.2 s , what is the magnitude of the induced emf in the loop while the current is changing?

A

$2.4 \mu \mathrm{~V}$

B

$32 \mu \mathrm{~V}$

C

$72 \mu \mathrm{~V}$

D

$4.8 \mu \mathrm{~V}$

3
TS EAMCET 2022 (Online) 18th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

An AC current is given by the expression, $I(t)=50 \sin (200 \pi t)$ in amperes. The frequency and rms value of the current, respectively are

A

$100 \mathrm{~Hz}, 50 \sqrt{2} \mathrm{~A}$

B

$100 \mathrm{~Hz}, 25 \sqrt{2} \mathrm{~A}$

C

$200 \mathrm{~Hz}, 50 \sqrt{2} \mathrm{~A}$

D

$200 \mathrm{~Hz}, 25 \sqrt{2} \mathrm{~A}$

4
TS EAMCET 2022 (Online) 18th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

An electromagnetic wave is propagating in vacuum along $-\hat{\mathbf{j}}$ direction. The magnetic field of the wave is given by $\mathbf{B}=\left(2 \times 10^{-8}\right) \cos \left[\pi \times 10^{15}\left(t+\frac{y}{c}\right)\right] \hat{\mathbf{k}} \mathrm{T}$. The electric field $\mathbf{E}$ of this wave is ( $c \equiv$ speed of light)

A

$E=4 \cos \left[\pi \times 10^{15}\left(t+\frac{y}{c}\right)\right] \hat{j} \mathrm{~V} / \mathrm{m}$

B

$E=6 \cos \left[\pi \times 10^{15}\left(t+\frac{y}{c}\right)\right] \hat{\mathrm{i}} \mathrm{V} / \mathrm{m}$

C
$\mathbf{E}=6 \cos \left[\pi \times 10^{15}\left(t-\frac{y}{c}\right)\right] \hat{\mathbf{j}} \mathrm{V} / \mathrm{m}$
D

$\mathbf{E}=4 \cos \left[\pi \times 10^{15}\left(t-\frac{y}{c}\right)\right] \hat{\mathbf{j}} \mathrm{V} / \mathrm{m}$

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