
A current carrying rectangular coil is placed in a uniform magnetic field. In which orientation will the coil not rotate?
A. The magnetic field is perpendicular to the plane of the coil.
B. The magnetic field is parallel to the plane of the coil.
C. The magnetic field is at ${45^ \circ }$ with the plane of the coil.
D. Always in any orientation.
Answer
163.2k+ views
Hint: To solve the above question, use the concept of torque experienced by a current carrying coil when it is placed in a uniform magnetic field. This torque is given by $\tau=NiAB \sin \theta=M~B~ \sin \theta$ , where A is the area of the loop and M is the magnetic dipole moment. The coil will not rotate when the value of the torque acting over it is 0. Use the formula to find out in which orientation will this torque be 0.
Formula used:
$\tau = NiAB\sin \theta $
Complete answer:
Torque on a current carrying loop placed in a uniform magnetic field is given by:
$\tau = NiAB\sin \theta $ … (1)
Where
$N$ is the number of loops in the coil,
$i$ is the current flowing through the coil,
$A$ is the area of the coil,
$B$ is the intensity of the magnetic field and
$\theta $ is the angle between the magnetic field and the axis of the coil.
This value of torque will be minimum when $\theta = 0$ or $\theta = \pi $ .
When $\theta = 0$ or $\theta = \pi $ , the value of torque on the current carrying rectangular coil will be 0 as $\sin \;0 = \sin \pi = 0$ .
As $\theta $ is the angle between the direction of the magnetic field and the axis of the coil, its value will be $0$ or $\pi $ , when the magnetic field is parallel or anti parallel to the direction of the axis.
Hence the plane of the coil should be perpendicular to the magnetic field. Thus, the correct option is A.
Note: For the coil to experience no torque (torque = 0), the cross product should give us a value equivalent to 0. This is only possible when the axis of the coil is parallel or anti parallel to the direction of the magnetic field.
Formula used:
$\tau = NiAB\sin \theta $
Complete answer:
Torque on a current carrying loop placed in a uniform magnetic field is given by:
$\tau = NiAB\sin \theta $ … (1)
Where
$N$ is the number of loops in the coil,
$i$ is the current flowing through the coil,
$A$ is the area of the coil,
$B$ is the intensity of the magnetic field and
$\theta $ is the angle between the magnetic field and the axis of the coil.
This value of torque will be minimum when $\theta = 0$ or $\theta = \pi $ .
When $\theta = 0$ or $\theta = \pi $ , the value of torque on the current carrying rectangular coil will be 0 as $\sin \;0 = \sin \pi = 0$ .
As $\theta $ is the angle between the direction of the magnetic field and the axis of the coil, its value will be $0$ or $\pi $ , when the magnetic field is parallel or anti parallel to the direction of the axis.
Hence the plane of the coil should be perpendicular to the magnetic field. Thus, the correct option is A.
Note: For the coil to experience no torque (torque = 0), the cross product should give us a value equivalent to 0. This is only possible when the axis of the coil is parallel or anti parallel to the direction of the magnetic field.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
