A wire carrying current 'I' along x axis has length 'L' and it is kept in a magnetic field B( i +2 j -2 k ) T. The magnitude of magnetic force acting on the wire is

Step-by-step solution

The magnetic force on a current-carrying wire is given by F = I( L B ) . Since the wire is along the x-axis, the length vector is L = L i . The magnetic field is given as B = B( i + 2 j - 2 k ) . Substituting these into the force formula: F = I[L i B( i + 2 j - 2 k )] F = ILB[ i i + 2( i j ) - 2( i k )] Using the cross product rules i i = 0 , i j = k , and i k = - j : F = ILB[0 + 2 k - 2(- j )] = 2ILB( j + k ) The magnitude of the magnetic force is: | F | = (2ILB)^2 + (2ILB)^2 = 8I^2L^2B^2 = 8 ILB Answer: 8 , ILB