A straight conductor of length l carrying current I is placed in a uniform magnetic field B so that the current makes an angle θ with the field. The magnitude of the force on the conductor is:

1. The force on a current-carrying conductor is $\mathbf{F} = I\,\mathbf{l} \times \mathbf{B}$. 2. The magnitude of a vector product is $|\mathbf{l} \times \mathbf{B}| = lB\sin\theta$, where $\theta$ is the angle between $\mathbf{l}$ and $\mathbf{B}$. 3. Therefore $F = IlB\sin\theta$. 4. The cos θ choice would wrongly give zero force when current is perpendicular to B, contradicting the maximum-force case. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.137_