1
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle of charge q and mass m is moving with a velocity $$ - v\widehat i$$ (v $$ \ne $$ 0) towards a large screen placed in the Y - Z plane at a distance d. If there is a magnetic field $$\overrightarrow B = {B_0}\widehat k$$ , the maximum value of v for which the particle will not hit the screen is :
A
$${{2qd{B_0}} \over m}$$
B
$${{qd{B_0}} \over {3m}}$$
C
$${{qd{B_0}} \over {2m}}$$
D
$${{qd{B_0}} \over {m}}$$
2
JEE Main 2020 (Online) 5th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
An iron rod of volume 10–3 m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :
A
5 $$ \times $$ 102 Am2
B
0.5 $$ \times $$ 102 Am2
C
500 $$ \times $$ 102 Am2
D
50 $$ \times $$ 102 Am2
3
JEE Main 2020 (Online) 5th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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 :
A
$${{2{\mu _0}{I^2}{a^2}} \over {\pi b}}$$
B
$${{{\mu _0}{I^2}{a^2}} \over {2\pi b}}$$
C
$${{{\mu _0}{I^2}{a^3}} \over {2\pi {b^2}}}$$
D
$${{2{\mu _0}{I^2}{a^3}} \over {\pi {b^2}}}$$
4
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
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