1
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
Out of Syllabus
At some location on earth, the horizontal component of earth’s magnetic field is 18 × 10–6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45o angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is -
A
3.6 $$\times$$ 10$$-$$5 N
B
1.8 $$\times$$ 10$$-$$5 N
C
1.3 $$\times$$ 10$$-$$5 N
D
6.5 $$\times$$ 10$$-$$5 N
2
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
A magnet of total magnetic moment 10-2 $${\widehat i}$$ A-m2 is placed in a time varying magnetic field, B$${\widehat i}$$ (cos $$\omega t$$) where B = 1 Tesla and $$\omega$$ = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is -
A
0.014 J
B
0.028 J
C
0.01 J
D
0.007 J
3
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The corecivity of the bar magnet is :
A
285 A/m
B
2600 A/m
C
520 A/m
D
1200 A/m
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
The $$B$$-$$H$$ curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solended with $$1000$$ turns/cm. The current that should be passed in the solended to demagnetise the ferromagnet completely is :

A
$$1$$ $$mA$$
B
$$2$$ $$mA$$
C
$$20\,\mu A$$
D
$$40\,\mu A$$
EXAM MAP
Medical
NEET