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 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:
(a)
3.6×10−5 N
(b)
6.5×10−5 N
(c)
1.3×10−5 N
(d)
1.8×10−5 N
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 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:
(a)
3.6×10−5 N
(b)
6.5×10−5 N
(c)
1.3×10−5 N
(d)
1.8×10−5 N
Question ID - 50256 | SaraNextGen Answer
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 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:
(a)
3.6×10−5 N
(b)
6.5×10−5 N
(c)
1.3×10−5 N
(d)
1.8×10−5 N
|
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 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is: |
|||
|
(a) |
3.6×10−5 N |
(b) |
6.5×10−5 N |
|
(c) |
1.3×10−5 N |
(d) |
1.8×10−5 N |
Bsin45°=F
sin45°
B