Describe the forces experienced by a current-carrying loop in a magnetic field and describe the net result of the forces

As a current carrying loop in an external magnetic field produces a magnetic field; this magnetic field will then interact with the external magnetic field and give a resultant force. As it is known by Faraday's Law that 'a current carrying conductor will feel a force within an external magnetic field' (otherwise known as the motor effect), we can deduce that this force will force the coil to move. Let us now examine the motor of which a current carrying conductor is situated between an external magnetic field. As the current moves from the positive terminal to the negative as shown in the diagram on the right, the conductor will feel a force, the direction of the force can be determined by use of the right hand rule which as shown in the diagram will force the conductor to move up on the left hand side and down on the right hand side. These forces result in a torque or a turning moment.

Due to the equation $F = ma$ we can see that any force acting on an object will accelerate that object, hence as torque is $Fd_\bot$ or $nBIAcos{\theta}$, torque will have direct bearing on acceleration. As $\Delta\tau$ $\alpha$ $\Delta F$ $\alpha$ $\Delta a$, the net result of these forces accelerates the coil in a rotational motion and provides turning force on the load. Hence the forces experienced by a current carrying loop will be a turning moment in the clockwise direction.

NOTE: the diagram shown electron movement it is a common HSC trick to show electron flow instead of conventional current to trick students, the right hand palm rule only works for conventional current.

HSC Style Question
5 marks
In the following diagram a motor is connected to a variable current: Identify the direction of rotation of the motor and describe the relationship between torque and acceleration, if magnetic field moves left to right.

As current flows through the coil in a clockwise direction as indicated by the circuit diagram, due to the motor effect the conductor will feel a force as the conductor induces a magnetic field to interact with the external magnetic field (Faraday). On the left hand side of the motor the conductor is perpendicular to the magnetic field lines and hence due to the right hand rule, it will feel a force downwards, conversely on the right hand side of the motor due to the right hand rule the direction of force is determined to be upwards. Since the remaining sides are parallel to the external magnetic field the fields will not interact and hence no force will be resultant. The net result of these forces will then provide a turning moment to the coil and due to the equation $\tau = nBIAcos{\theta} = Fd_\bot$ force is acting on a distance with applied force perpendicular to the resultant force hence $F \alpha \tau$. Due to the equation $F = ma$, any force acting upon an object will cause the object to accelerate and thus force (is directly proportional to) acceleration. Thus torque (is directly proportional to) acceleration and the coil will accelerate in the diagram.