Choose the correct option. A conducting thick copper rod of length 1 m carries a current of 15A and is located on the Earth's equator. There the magnetic flux lines of Earth's magnetic field are horizontal, with the field of 1.3 x 10-4T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east are ______.

Question

KollegeApply · Accepted Answer

A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth's equator. There the magnetic flux lines of Earth's magnetic field are horizontal, with the field of 1.3 $\times$ 10-4T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east are 20 $\times$ 10-4N, upward. Explanation: Express the relation for the force acting on a current-carrying wire due to a perpendicular magnetic field. vecF = i * L vec$\times$ B Here, F is force, B is the perpendicular magnetic field, i is current and L is the length of the conductor. Substitute 1.3 $\times$ 10−4T for B, 1 m for L and 15A for I in the equation to find the magnitude of the force. F = `(1.3 xx 10^-4T) xx (15A) xx (1m) F = 19.5 xx 10^-4N F ≈ 20 xx 10^-4N` The direction of the force is found by using Fleming’s left-hand rule where the current corresponds to the middle finger, the field corresponds to the pointing finger and the force corresponds to the thumb. Hence the force is upwards.

AChoose Your Answer