JEE PYQ: Magnetism And Matter Question 19
Question 19 - 2019 (10 Jan 2019 Shift 2)
At some location on earth the horizontal component of earth’s magnetic field is $18 \times 10^{-6}$ 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 $45°$ angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:
(1) $3.6 \times 10^{-5}$ N
(2) $1.8 \times 10^{-5}$ N
(3) $1.3 \times 10^{-5}$ N
(4) $6.5 \times 10^{-5}$ N
Show Answer
Answer: (4) $6.5 \times 10^{-5}$ N
Solution
$F = 2MB = 2 \times 1.8 \times 18 \times 10^{-6} = 6.5 \times 10^{-5}$ N
BETA
