JEE Main 2019 (Online) 12th April Morning Slot | Magnetics Question 174 | Physics

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 $$\pi $$ rad s^–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10^–9 T, then the charge carried by the ring is close to ($$\mu $$₀ = 4$$\pi $$ × 10^–7 N/A² ).

7 × 10^–6 C

4 × 10^–5 C

2 × 10^–6 C

3 × 10^–5 C

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1 cm s^-1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 $$\Omega $$, the current in the loop at that instant will be close to : JEE Main 2019 (Online) 12th April Morning Slot Physics - Magnetics Question 176 English

JEE Main 2019 (Online) 12th April Morning Slot Physics - Magnetics Question 176 English

115 $$\mu $$A

170 $$\mu $$A

60 $$\mu $$A

150 $$\mu $$A

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

An electromagnetic wave is represented by the electric field $$\overrightarrow E = {E_0}\widehat n\sin \left[ {\omega t + \left( {6y - 8z} \right)} \right]$$ . Taking unit vectors in x, y and z directions to be $$\widehat i,\widehat j,\widehat k$$ , the direction of propagation $$\widehat s$$, is :

$$\widehat s = {{3\widehat i - 4\widehat j} \over 5}$$

$$\widehat s = {{ - 4\widehat k + 3\widehat j} \over 5}$$

$$\widehat s = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)$$

$$\widehat s = {{4\widehat j - 3\widehat k} \over 5}$$

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45^o, and 40 times per minute where the dip is 30^o. If B₁ and B₂ are respectively the total magnetic field due to the earth at the two places, then the ratio $${{{B_1}} \over {{B_2}}}$$ is best given by :

1.8

2.2

0.7

3.6

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