JEE Main 2020 (Online) 4th September Evening Slot | Magnetics Question 140 | Physics

JEE Main 2020 (Online) 4th September Evening Slot

MCQ (Single Correct Answer)

-1

A circular coil has moment of inertia 0.8 kg m² around any diameter and is carrying current to produce a magnetic moment of 20 Am² . The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical,it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60^o will be:

10 $$\pi $$ rad s^–1

20 $$\pi $$ rad s^–1

$$10{\left( 3 \right)^{1/4}}$$ rad s^–1

20 rad s^–1

JEE Main 2020 (Online) 4th September Evening Slot

MCQ (Single Correct Answer)

-1

The electric field of a plane electromagnetic wave is given by
$$\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
Its magnetic field will be given by :

$${{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$

$${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)$$

$${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)$$

$${{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$

JEE Main 2020 (Online) 4th September Evening Slot

MCQ (Single Correct Answer)

-1

A paramagnetic sample shows a net magnetisation of 6 A/m when it is placed in an external magnetic field of 0.4 T at a temperature of 4 K. When the sample is placed in an external magnetic field of 0.3 T at a temperature of 24 K, then the magnetisation will be:

4 A/m

1 A/m

0.75 A/m

2.25 A/m

JEE Main 2020 (Online) 4th September Morning Slot

MCQ (Single Correct Answer)

-1

A small bar magnet placed with its axis
at 30^o with an external field of 0.06 T
experiences a torque of 0.018 Nm. The
minimum work required to rotate it from its
stable to unstable equilibrium position is :

6.4 $$ \times $$ 10^-2 J

9.2 $$ \times $$ 10^-3 J

7.2 $$ \times $$ 10^-2 J

11.7 $$ \times $$ 10^-3 J

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