1
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-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 60o will be:
A
10 $$\pi $$ rad s–1
B
20 $$\pi $$ rad s–1
C
$$10{\left( 3 \right)^{1/4}}$$ rad s–1
D
20 rad s–1
2
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-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 :
A
$${{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
B
$${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
C
$${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)$$
D
$${{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
3
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-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:
A
4 A/m
B
1 A/m
C
0.75 A/m
D
2.25 A/m
4
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
A small bar magnet placed with its axis
at 30o 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 :
A
6.4 $$ \times $$ 10-2 J
B
9.2 $$ \times $$ 10-3 J
C
7.2 $$ \times $$ 10-2 J
D
11.7 $$ \times $$ 10-3 J
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