JEE Main 2020 (Online) 5th September Morning Slot | Magnetics Question 137 | Physics

JEE Main 2020 (Online) 5th September Morning Slot

MCQ (Single Correct Answer)

-1

An electron is constrained to move along the y-axis with a speed of 0.1 c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is
$$\overrightarrow E = 30\widehat j\sin \left( {1.5 \times {{10}^7}t - 5 \times {{10}^{ - 2}}x} \right)$$ V/m.
The maximum magnetic force experienced by the electron will be :
(given c = 3 $$ \times $$ 10⁸ ms^–1 and electron charge = 1.6 $$ \times $$ 10^–19 C)

4.8 $$ \times $$ 10^–19 N

2.4 $$ \times $$ 10^–18 N

3.2 $$ \times $$ 10^–18 N

1.6 $$ \times $$ 10^–18 N

JEE Main 2020 (Online) 5th September Morning Slot

MCQ (Single Correct Answer)

-1

A square loop of side 2$$a$$, and carrying current I, is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z-axis and passing through the point (0, b, 0), (b >> a). The magnitude of the torque on the loop about zaxis is given by :

$${{2{\mu _0}{I^2}{a^2}} \over {\pi b}}$$

$${{{\mu _0}{I^2}{a^2}} \over {2\pi b}}$$

$${{{\mu _0}{I^2}{a^3}} \over {2\pi {b^2}}}$$

$${{2{\mu _0}{I^2}{a^3}} \over {\pi {b^2}}}$$

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 :

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

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

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

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

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