1 , C charge moving with velocity v = ( i - 2 j + 3 k ) m/s in the region of magnetic field B = (2 i + 3 j - 5 k ) T. The magnitude of force acting on it is 10⁻⁶ N. The value of is _______.

Step-by-step solution

The magnetic force acting on a moving charge is given by F = q( v B ) . Given: q = 1 , C = 10⁻⁶ C v = i - 2 j + 3 k B = 2 i + 3 j - 5 k Calculating the cross product v B : v B = vmatrix i & j & k 1 & -2 & 3 2 & 3 & -5 vmatrix v B = i (10 - 9) - j (-5 - 6) + k (3 + 4) v B = i + 11 j + 7 k The magnitude of v B is: | v B | = 1^2 + 11^2 + 7^2 = 1 + 121 + 49 = 171 The magnitude of the force is: | F | = q | v B | = 10⁻⁶ 171 = 171 10⁻⁶ N Comparing this with the given expression 10⁻⁶ N , we get: = 171 Answer: 171