Question

A current-carrying coil is placed in a magnetic field. Which effect does the coil experience?

A) A change in shape
B) A change in weight
C) A resultant force
D) A turning effect

Answer 1

Hint: When a moving charge is in a magnetic field it experiences a force $\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$. Here F = force, q = Charge; v = Velocity, B = magnetic field. In a current carrying wire there are many charges flowing through a conductor and they are in uniform magnetic field, so they will experience a force.

Complete answer:
Force on one e = $e\left( {{{\overrightarrow V }_d} \times \overrightarrow B } \right) = e{v_d}B\sin \theta $;
Force on all e in elemental region =$Ne{v_d}B\sin \theta $;
N is the number of electrons passing through an elemental area of a certain length:
$N = nAd\operatorname{l} $;…(n = number of free electrons passing through area A of length dl)
$dF = nAd\operatorname{l} e{v_d}B\sin \theta $;
Now, to calculate the force on the whole wire we have to integrate on both the sides:
$\int {dF} = \int {nAd\operatorname{l} e{v_d}B\sin \theta } $;
$ \Rightarrow F = neA{v_d}B\sin \theta \int {d\operatorname{l} } $;
Integrate “dl” from zero to l:
$ \Rightarrow F = neA{v_d}B\sin \theta \int\limits_0^l {dl} $;
$ \Rightarrow F = neA{v_d}B\sin \theta l$;
Now, we know current in terms of drift velocity “${v_d}$”.
$i = {v_d}enA$;
So,
$F = iBl\sin \theta $;
Now, to calculate the net force on the current carrying wire due to uniform magnetic field we need to find the angle between the magnetic field and the current carrying wire, here there can be two possibilities one is the current carrying wire is parallel to magnetic field and another it can be perpendicular to the magnetic field.
Here, for parallel cases the angle between magnetic field and current carrying wire is zero.
$F = iBl\sin 0$;
$\Rightarrow F = 0$;
For perpendicular case the angle between them would be $90^\circ $:
$F = iBl\sin 90$;
$\Rightarrow F = iBl$;
Here the force will be applied on the two ends of the wire so:
$\Rightarrow {F_1} = iBl$;
$\Rightarrow {F_2} = iBl$;
The above two forces are equal and opposite to each other and hence they will cancel out but they will produce a torque on the loop which will try to rotate it anticlockwise.

Final answer is option (D) is correct. Therefore, the coil experiences a turning effect.

Note: Here, contrary to our intuition that there will be a net force acting on the wire due to the presence of a magnetic field the forces indeed are there but they cancel each other out and hence there would be no net force.