Question

A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times {10^{ - 4}}{m^2}$carries a current of 2.0A. It is suspended through its center and perpendicular its length, allowing it to turn in a horizontal plane in a uniform magnetic field$5 \times {10^{ - 2}}$tesla making an angle of ${30^ \circ }$with the axis of the solenoid. The torque on the solenoid will be
A. $3 \times {10^{ - 3}}Nm$
B. $1.7 \times {10^{ - 3}}Nm$
C. $1.5 \times {10^{ - 2}}Nm$
D. $3 \times {10^{ - 2}}Nm$

Answer 1

Hint: The solenoid can be here treated as a current-carrying loop. The torque acting on a current-carrying loop is given by the cross product of the magnetic moment and magnetic field. And the magnetic moment is given as the product of the number of turns in the coil, the current passing through the coil, and it’s the cross-sectional area. Using these equations mathematically, we can get the required answer.

Formula Used:
Torque on a solenoid, $\tau = \vec M \times \vec B = MB\sin \theta $
where
$\eqalign{
  & M{\text{ is the magnetic moment,}} \cr
  & B{\text{ is the magnetic field,}} \cr
  & \theta {\text{ is the angle between the magnetic moment vector and magnetic field vector}}{\text{.}} \cr} $
Magnetic moment, $M = NIA$
$\eqalign{
  & N{\text{ is the number of turns in the solenoid,}} \cr
  & I{\text{ is the current through the solenoid,}} \cr
  & A{\text{ is the area of cross - section of the solenoid}}{\text{.}} \cr} $

Complete step by step answer:
A solenoid means an insulated copper wire wound closely in the form of a helix. The word solenoid comes from the Greek word meaning channel. A solenoid can thus be considered as a current carrying loop. The magnetic field when acts on a current carrying loop extends a torque on it. This torque can be given mathematically as:
$\tau = \vec M \times \vec B = MB\sin \theta \cdots \cdots \cdots \cdots \left( 1 \right)$
where
$\eqalign{
  & M{\text{ is the magnetic moment,}} \cr
  & B{\text{ is the magnetic field,}} \cr
  & \theta {\text{ is the angle between the magnetic moment vector and magnetic field vector}}{\text{.}} \cr} $
Additionally, the magnetic moment of a solenoid i.e., the strength and direction of the solenoid for the applied magnetic field is given by:
$M = NIA \cdots \cdots \cdots \cdots \left( 2 \right)$
where
$\eqalign{
  & N{\text{ is the number of turns in the solenoid,}} \cr
  & I{\text{ is the current through the solenoid,}} \cr
  & A{\text{ is the area of cross - section of the solenoid}}{\text{.}} \cr} $

Given:
Number of turns in the solenoid, $N = 2000$turns
Area of cross-section, $A = 1.5 \times {10^{ - 4}}{m^2}$
Current, $I = 2.0A$
Magnetic field, $B = 5 \times {10^{ - 2}}$tesla
Angle, $\theta = {30^ \circ }$

Substituting values in equation (2), we get:
$\eqalign{
  & M = NIA \cr
  & \Rightarrow M = 2000 \times 2 \times 1.5 \times {10^{ - 4}} \cr
  & \Rightarrow M = 0.6A{m^2} \cr} $
Now, substituting the found value of magnetic moment in equation (1), we get:
$\eqalign{
  & \tau = \vec M \times \vec B = MB\sin \theta \cr
  & \Rightarrow \tau = 0.6 \times 5 \times \dfrac{1}{2} \cr
  & \therefore \tau = 1.5 \times {10^{ - 2}}Nm \cr} $
Therefore, the correct option is C i.e., the torque acting on the solenoid is .$1.5 \times {10^{ - 2}}Nm$.

Note: The magnetic field inside a toroidal solenoid is independent of its radius and depends only on the current through the solenoid and the number of turns per unit length. The field inside the toroid has a constant magnitude and tangential direction at every point.