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Torque On Current Loop And Magnetic Moment Derivation MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Torque on a rectangular current loop in a uniform magnetic field, Circular current loop as magnetic dipole is considered one of the most asked concept.

  • 58 Questions around this concept.

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 A magnetic dipole in a constant magnetic field has :

A closely wound solenoid of 2000 turns and area of cross-section 1.5\times 10^{-4}m^{2} carries a current of 2.0 A. It is suspended through its centre and perpendicular to 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^{o} with the axis of the solenoid. The torque on the solenoid will be

A bar magnet having a magnetic moment  2\times 10^{4}JT^{-1} is free to rotate in a horizontal plane. A horizontal magnetic field B=6\times 10^{-4}T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60^{o} from the field is

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A uniform magnetic field B of 0.3 T is along the positive Z-direction.  A rectangular loop (abcd) of sides 10 cm×5 cm carries a current I of 12 A.  Out of the following different orientations which one corresponds to stable equilibrium ?

 

 A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below :

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium ?

The magnetic susceptibility is negative for:

Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole momentm . Which configuration has highest net magnetic dipole moment?

(a)     (b) 

(c)     (d) 

 

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Concepts Covered - 2

Torque on a rectangular current loop in a uniform magnetic field

Torque on a rectangular current loop in a uniform magnetic field - 

As we have studied that the electric dipole in a uniform electric field it will experience a torque similarly if we place a rectangular loop carrying a steady current i and placed in a uniform magnetic field experiences a torque. It does not experience a net force.

Let us consider a case when the rectangular loop is placed such that the uniform magnetic field B is in the plane of the loop. This is illustrated in the given figure.The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F1 on it which is directed into the plane of the loop. Its magnitude is,

                                                                                              F_{1}=I b B

                                                                   

Similarly it exerts a force F2 on the arm CD and F2 is directed out of the plane of the paper.

                                                                                        F_{2}=I b B=F_{1}

Thus, the net force on the loop is zero. But these two forces are actingat a distance 'a' between them. This torque on the loop due to the pair of forces F1 and F2 . From the figure given below shows that the torque on the loop tends to rotate it anti-clockwise. This torque is (in magnitude),

                                                                                 

                                                                         

                                                                                 \begin{aligned} \tau &=F_{1} \frac{a}{2}+F_{2} \frac{a}{2} \\ &=I b B \frac{a}{2}+I b B \frac{a}{2}=I(a b) B \\ &=I A B \end{aligned}

                                                                           where A = ab is the area of the rectangle.

Now we will discuss the case when the plane of the loop is making an angle \theta with magnetic field. In the previous case we have considered  \theta = \frac{\pi}{2} , but this is now a general case. 

                                                             

Here again you can see that the forces on arms AB and CD are F1 and F2 - 

                                                                                    F_{1}=F_{2}=I b B

Then the torque will be the

                                                                        \begin{aligned} \tau &=F_{1} \frac{a}{2} \sin \theta+F_{2} \frac{a}{2} \sin \theta \\ &=\operatorname{Iab} B \sin \theta \\ &=I A B \sin \theta \end{aligned}

From the above equations we can see that the torques can be expressed as vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as,

                                                                                            m = I A

If the coil has N turns then the magnetic moment formula becomes - 

                                                                                         m = N I A

And its direction is defined by the direction of Area vector.

So, Torque equation can be written as, 

                                                                                         \tau=\mathbf{m} \times \mathbf{B}

 

 

Circular current loop as magnetic dipole

Circular current loop as magnetic dipole-

From the previous concept, we have studied the magnetic field due a current (I) carrying circular wire (Radius = R) on its axis at distance 'x' is -

                                                                          B=\frac{\mu_{0} I R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}}

If x>>>R, then R will become negligible and the equation become - 

                                                                                 B=\frac{\mu_{0} R^{2}}{2 x^{3}}

Now, as the area of this loop is \pi R^2, so the equation become - 

                                                                                 B=\frac{\mu_{0} I A}{2 \pi x^{3}}

As, m = I A So, 

                                                                                 \begin{aligned} \mathbf{B} & = \frac{\mu_{0} \mathbf{m}}{2 \pi x^{3}} \\ \\ &=\frac{\mu_{0}}{4 \pi} \frac{2 \mathbf{m}}{x^{3}} \end{aligned}

 

The expression shown above is very similar to an expression obtained earlier for the electric field of a dipole.

The similarity may be seen if we substitute,

                                                                   \begin{array}{l}{\mu_{0} \rightarrow 1 / \varepsilon_{0}} \\ \\ {\mathbf{m} \rightarrow \mathbf{p}_{e} \text { (electrostatic dipole) }} \\ \\ {\mathbf{B} \rightarrow \mathbf{E} \text { (electrostatic field) }}\end{array}

So the equation become ,

                                                                                 \mathbf{E}=\frac{2 \mathbf{p}_{e}}{4 \pi \varepsilon_{0} x^{3}}

So, We can say from the above analogy that the circular current loop can act as a magnetic dipole. The direction of the magnetic moment can be obtain as - 

                                                              

But there is a fundamental difference: an electric dipole is built up of two elementary units — the charges (or electric monopoles). In magnetism, a magnetic dipole (or a current loop) is the most elementary element. The equivalent of electric charges, i.e., magnetic monopoles, are not known to exist.

 

Study it with Videos

Torque on a rectangular current loop in a uniform magnetic field
Circular current loop as magnetic dipole

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