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 :

Question

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

Tardigrade Admin · Accepted Answer

μ B sin45°=F (ℓ/2) sin45° F =2μ B

$3.6 \times 1 0^{- 5} N$

$6.5 \times 1 0^{- 5} N$

$1.3 \times 1 0^{- 5} N$

$1.8 \times 1 0^{- 5} N$

$μ B sin 4 5^{\circ} = F \frac{ℓ}{2} sin 4 5^{\circ}$
$F = 2 μ B$

3.6×10−5N

6.5×10−5N

1.3×10−5N

1.8×10−5N

μBsin45∘=F2ℓ​sin45∘ F=2μB

$3.6 \times 1 0^{- 5} N$

$6.5 \times 1 0^{- 5} N$

$1.3 \times 1 0^{- 5} N$

$1.8 \times 1 0^{- 5} N$

$μ B sin 4 5^{\circ} = F \frac{ℓ}{2} sin 4 5^{\circ}$
$F = 2 μ B$