Question

The magnetic field goes to the left of the screen as shown. While the constant current travels at an angle $\theta$ to the upper right corner. Determine the force per unit length of wire on this wire.
$\text{A.}\quad \dfrac{IB}{cos\theta}$
$\text{B.}\quad \dfrac{IB}{sin\theta}$
$\text{C.}\quad {IB}{sin\theta}$
$\text{D.}\quad {IB}{cos\theta}$

Answer 1

Hint: Magnetic field is the space around the magnet where the effect of magnet can be felt by another magnet or iron piece. Magnetic fields can also be produced by a moving charge whose intensity can be determined by the velocity and magnitude of charge. The S.I unit of magnetic field is Tesla (T) whereas the C.G.S unit is Gauss (G).
Formula used:
$\vec F = i(\vec l \times \vec B)$

Complete answer:
Since the moving charge produces a magnetic field, hence if an external magnetic field is applied, the moving charge will interact with the field and hence it experiences a force or magnetic force. As current is also a movement of charge, so a current carrying wire will also experience a force. This force is given by $\vec F = i(\vec l \times \vec B) \ or\ Bil\ sin\theta$where $\theta$ is the angle between ‘l’ and ‘B’.
Now, $F = Bil\ sin\theta$
$\implies F = BI L sin(\pi -\theta) = BIL sin\theta$
Now, force per unit length$= \dfrac{F}{L} = BI sin\theta$

So, the correct answer is “Option C”.

Note:
Students are advised to revise the properties of the cross product and the direction it gives. The direction of the force acting on a current carrying wire is perpendicular to both magnetic field and length of the wire. This direction could also be predicted using left hand rule. Also in the above question, we have to take the angle between length and magnetic field. In this question, we have this angle as $(\pi - \theta)$ and not simply $\theta$. We can also see that the force will be maximum if the magnetic field is perpendicular to the length of the wire and minimum force (i.e. zero) if the length of wire is kept in the direction of magnetic field.