At some location 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 ° angles with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:

Step-by-step solution

The horizontal and vertical components of earth's magnetic field ( B H and B V ) are related as B V B H = tan θ Here, θ = 45 ° and B H = 18 × 10 - 6   T ⇒ B V = B H tan 45 ° ⇒ B V = B H = 18 × 10 - 6   T ∵   tan 45 ° = 1 Now, when the external force F is applied, to keep the needle stays in horizontal position is shown below, Taking torque at point P , we get m B V × 2 l = F l ∴   F = 2 × m B V Substituting the given values