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 $1.8 A m$ 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

see full answer

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

$3.6 \times 10^{- 5} N$

$1.3 \times 10^{- 5} N$

$1.8 \times 10^{- 5} N$

$6.5 \times 10^{- 5} N$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$B_{H} = 18 \times 10^{- 6} T, m = 1.8 A m, l = 0.12 m$

$|τ_{F}| = |τ_{F B}| \Rightarrow F \times \frac{l}{2} \sin 45 ° = μ B \sin 45 °$

$F = 2 μ B / l = 2 m B = 2 \times 1.8 \times 18 \times 10^{- 6} = 6.48 \times 10^{- 5} N$

Watch 3-min video & get full concept clarity