A straight wire of mass 200 g and length 5 m carries a current of 2 A . It is oriented horizontally along the vector L = 3 i + 4 j (in meters). The wire is suspended in mid-air, balancing gravity which acts in the - k direction, by a uniform magnetic field of the form B = B₀ i + B₀ j + B_z k . What is the required magnetic field vector B ? (Take g = 10 m/s ^2 )

Step-by-step solution

The gravitational force on the wire is W = -mg k = -0.2 10 k = -2 k N . For the wire to be in mechanical equilibrium, the magnetic force must balance the weight: F _m = +2 k N The magnetic force on a current-carrying wire is given by F _m = I( L B ) . Here, I = 2 A and L = 3 i + 4 j . Calculating the cross product: F _m = 2 [ (3 i + 4 j ) (B₀ i + B₀ j + B_z k ) ] F _m = 2 [ 3B₀( i i ) + 3B₀( i j ) + 3B_z( i k ) + 4B₀( j i ) + 4B₀( j j ) + 4B_z( j k ) ] F _m = 2 [ 0 + 3B₀ k - 3B_z j - 4B₀ k + 0 + 4B_z i ] F _m = 8B_