Welcome to this series on Objective Physics Question Bank for NTA NEET, JEE Mains & WBJEE aspirants! The present section contains a list of the most important and conceptual physics MCQs on the topic

Special care has been taken in selection of problems to ensure that they fall under the syllabus prescribed by the authorities and of the right difficulty level as demanded by the said exams. In fact a fair portion of questions are directly taken from the archives of these exams. This makes this series even more useful and relevant for the students to be used as a DPP (daily practice problems) or regular assignments.

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**Magnetic Moment of a Current Carrying Loop/Coil & Torque on it when placed in magnetic field**of the chapter Magnetic Effect of Electric Current. This topic includes following subtopics/contents:- Magnetic moment associated with a current carrying coil
- Magnetic moment associated with a rotating charged body
- Torque on a current coil placed in uniform magnetic field
- Potential Energy of a current coil associated with external magnetic field.

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**OBJECTIVE QUESTION BANK**

Out of the four different choices following each question, only one is the correct (or the most appropriate) which you have to identify and mark.

2. A current carrying loop behaves like tiny magnet. If area of the loop is A and magnetic moment is M, then current through the loop is

4. The magnetic moment (M) of a revolving electron around the nucleus varies with principle quantum number 'n' as:

5. The magnetic moment of a circular orbit of radius r carrying a charge

8. A current carrying circular loop of radius R produces a magnetic field B at its centre. The magnetic moment of the loop is

10. A current carrying coil is oriented in a constant uniform magnetic field B. Torque is maximum on this coil when plane of coil is

11. Magnetic field is parallel to the plane of a current coil, then torque will be

12. A current coil is in equilibrium in a uniform magnetic field. Nature of the equilibrium

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13. A circular coil of 100 turns and having a radius of 0.05 m carries a current of 0.1 A. The plane of coil is initially at right angles to magnetic field. What is the work required to turn the coil in an external field of 1.5 T through 180° about an axis perpendicular to the magnetic field?
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Also from Bhor's postulates, angular momentum of the revolving electron = nh/2Ï€. Or, mvr = nh/2Ï€, or vr ∝ n.

Hence, M ∝ vr ∝ n.

Magnetic moment, M = iA = (qv/2Ï€r)(Ï€r²) = qvr/2.

∴ M = iÏ€R²/2 = iÏ€L²/2(Ï€+2)².

First write the elemental magnetic moment associated with this small part and then integrate the expression for the whole length of the rod. This will give you the net magnetic moment of the system. You may visit my

Another way is to directly use the concept of gyromagnetic ratio (for the given system) as follows:

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Potential energy of the current coil associated with external magnetic field is given by U = -

We observe that Î¸ = 0° corresponds to the minimum potential energy and Î¸ = 180° corresponds to the maximum potential energy. Hence, the

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Work required to turn the coil from initial position to final = U₂ - U₁ = MB - (-MB) = 2MB.

Also, M = niA = niÏ€r² = (100)(0.1)(3.14)(0.05)² = 0.0785 A-m² and B = 1.5 T (given). ∴ Work, W = 2MB = 2(0.0785)(1.5) = 0.2355 J.

1. The SI unit of magnetic moment is

2. A current carrying loop behaves like tiny magnet. If area of the loop is A and magnetic moment is M, then current through the loop is

3. If number of turns, area and current through the coil is given by n, A and i respectively then its magnetic moment will be given by

4. The magnetic moment (M) of a revolving electron around the nucleus varies with principle quantum number 'n' as:

5. The magnetic moment of a circular orbit of radius r carrying a charge

*q*and rotating with velocity

*v*is given by:

*²/2*

*²*

*/2*

8. A current carrying circular loop of radius R produces a magnetic field B at its centre. The magnetic moment of the loop is

9. A current loop of area 0.01 m² and carrying a current of 10 A is held perpendicular to a magnetic field of 0.1 T, the torque in N-m acting on the loop is:

10. A current carrying coil is oriented in a constant uniform magnetic field B. Torque is maximum on this coil when plane of coil is

11. Magnetic field is parallel to the plane of a current coil, then torque will be

12. A current coil is in equilibrium in a uniform magnetic field. Nature of the equilibrium

13. A circular coil of 100 turns and having a radius of 0.05 m carries a current of 0.1 A. The plane of coil is initially at right angles to magnetic field. What is the work required to turn the coil in an external field of 1.5 T through 180° about an axis perpendicular to the magnetic field?

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**ANSWERS WITH HINTS**

**1. B Magnetic moment of a planer current loop is given by M = iA where i is the current and A stands for the loop area. From this, SI unit of M = A-m².**

**2. A Magnetic moment of a planer current loop is given by M = iA where i is the current and A stands for the loop area. This implies, i = M/A.**

**3. A Magnetic moment of a planer coil is defined as M =niA where n = number of turns, i = current and A = area enclosed.**

**4. A**If electron revolves in a circle of radius r with speed v then, average current in the loop, i = e/T = e/(2Ï€r/v) = ev/2Ï€r. Magnetic moment of the orbit, M = iA = (ev/2Ï€r)(Ï€r²) = evr/2. Or, M ∝ vr.

Also from Bhor's postulates, angular momentum of the revolving electron = nh/2Ï€. Or, mvr = nh/2Ï€, or vr ∝ n.

Hence, M ∝ vr ∝ n.

**5. B**Effective or average current in the orbit = q/T where T is the time period of rotation. i = q/T = q/(2Ï€r/v) = qv/2Ï€r.

Magnetic moment, M = iA = (qv/2Ï€r)(Ï€r²) = qvr/2.

**6. D**If R be the radius then, Magnetic moment = iA = i(Ï€R²/2). But R is not given.Total length (perimeter) of the wire = L = (Ï€+2)R. Or, R = L/(Ï€+2).

∴ M = iÏ€R²/2 = iÏ€L²/2(Ï€+2)².

**7. A**The rod can be thought of being constituted of large number of small sections/elements. These charge elements are performing uniform circular motions of varying radii with common angular velocity Ï‰. Effectively, they are constituting current loops and therefore giving rise to certain magnetic moments. Consider one such small part/element of the rod, located at a distance x from the axis of rotation and having a length dx and charge dq.

First write the elemental magnetic moment associated with this small part and then integrate the expression for the whole length of the rod. This will give you the net magnetic moment of the system. You may visit my

**this**article on quora for more details of this method.

Another way is to directly use the concept of gyromagnetic ratio (for the given system) as follows:

**8. B**Magnetic field at the center of circular current loop: B = Î¼â‚’i/2R (Using Biot-Savart Law).

∴ i = 2BR/Î¼â‚’. Magnetic moment, M = iA = (2BR/Î¼â‚’) (Ï€R²) = 2Ï€BR³/Î¼â‚’.

**9. D**Magnetic torque acting on the loop has a magnitude given by, Ï„ = MBsinÎ¸ where Î¸ is the angle between

**M**and

**B.**As the loop is held perpendicular to magnetic field,

**M**is either parallel or anti-parallel to

**B**depending on the direction of current in the loop. This means Î¸ is either 0° or 180°. In both the cases, Ï„ = 0.

**10. B**Magnetic torque on the coil has magnitude, Ï„ = MBsinÎ¸. For a given current coil and a given magnetic field, Ï„ will be maximum when sinÎ¸ is maximum i.e. sinÎ¸ = 1 or Î¸ = 90°. This means, angle between

**M**and

**B**vectors should be 90°. For this to happen, the current coil must be held parallel to

**B**.

**11. A**Magnetic field is parallel to the plane of a current coil. This implies magnetic field B is perpendicular to the magnetic moment M (since magnetic moment is always directed normal to the plane of coil). Thus, Î¸ = 90° in Ï„ = MBsinÎ¸ and therefore the torque is maximum (as sinÎ¸ is maximum at Î¸ = 90°).

**12. D**Under equilibrium, torque on the coil is zero i.e. Ï„ = MBsinÎ¸ = 0, or sinÎ¸ = 0. So, the coil is in equilibrium for two values of Î¸ viz. Î¸ = 0° and Î¸ = 180°. Let's check out the potential energy at these values of Î¸.

Potential energy of the current coil associated with external magnetic field is given by U = -

**M.B**= - MBcosÎ¸. At Î¸ = 0°, U = - MBcos0° = - MB. Similarly at Î¸ = 180°, U = - MBcos180° = +MB.

We observe that Î¸ = 0° corresponds to the minimum potential energy and Î¸ = 180° corresponds to the maximum potential energy. Hence, the

*equilibrium at*

*Î¸ = 0° is stable and that at*

*Î¸ = 180° is unstable*.

**13. C**Potential energy of the current coil associated with external magnetic field is given by U = -

**M.B**= - MBcosÎ¸. Initially Î¸ = 0° and U₁ = -MBcos0° = - MB. Finally, Î¸ = 180° and U₂ = -MBcos180° = MB.

Work required to turn the coil from initial position to final = U₂ - U₁ = MB - (-MB) = 2MB.

Also, M = niA = niÏ€r² = (100)(0.1)(3.14)(0.05)² = 0.0785 A-m² and B = 1.5 T (given). ∴ Work, W = 2MB = 2(0.0785)(1.5) = 0.2355 J.

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