
A bar magnet of magnetic moment $200A - {m^2}$ is suspended in a magnetic field of intensity $0.25N/A - m$ . The couple required to deflect it through ${30^ \circ }$ is
A. $50N - m$
B. $25N - m$
C. $20N - m$
D. $15N - m$
Answer
133.8k+ views
Hint:
When a bar magnet is suspended in a magnetic field of intensity, the two poles of the magnet experience a force. These experienced forces are equal in magnitude and opposite in direction. Thereby constituting a couple of forces that produce a torque. By applying the formula of torque we can calculate the desired deflection.
Formula used:
The formula used to calculate the couple in the solution is: -
$\tau = \overrightarrow M \times \overrightarrow B = MB\sin \theta $
Here $\tau =$ Net force on magnet or Torque
$M=$ Magnetic moment
$B=$ Magnetic field of intensity
$\theta =$ Angle between the magnetic field and magnetic axes of the bar magnet
Complete step by step solution:
A bar magnet of magnetic moment $M = 200A - {m^2}$ is suspended in a magnetic field of intensity, $B = 0.25N/A - m$ (given)
Now we know that the formula for the torque acting on a magnet in a uniform magnetic field is given as:
$\tau = \overrightarrow M \times \overrightarrow B = MB\sin \theta $ … (1)
This magnet experiences a couple (or torque) in the magnetic field. Now to find out the couple required to deflect it through $\theta = {30^ \circ }$ we will take help from equation (1).
$ \Rightarrow \tau = MB\sin \theta $
Substituting all the required values from the question in the above expression, we get
$ \Rightarrow \tau = (200) \times 0.25 \times \sin \left( {{{30}^ \circ }} \right)$
$ \Rightarrow \tau = 50 \times 0.5 = 25N - m$ $\left( {\therefore \sin {{30}^ \circ } = \dfrac{1}{2} = 0.5} \right)$
Thus, the couple required to deflect a bar magnet through ${30^ \circ }$ is $25N - m$.
Hence, the correct option is (B) $25N - m$ .
Therefore, the correct option is B.
Note:
The alpha particles, beta particles and protons are charged particles, thereby being deflected by magnetic fields. Only gamma rays which are actually neutron particles are not deflected by magnetic fields as these particles have no charge. Neutron does not contain any electrical charge particles.
When a bar magnet is suspended in a magnetic field of intensity, the two poles of the magnet experience a force. These experienced forces are equal in magnitude and opposite in direction. Thereby constituting a couple of forces that produce a torque. By applying the formula of torque we can calculate the desired deflection.
Formula used:
The formula used to calculate the couple in the solution is: -
$\tau = \overrightarrow M \times \overrightarrow B = MB\sin \theta $
Here $\tau =$ Net force on magnet or Torque
$M=$ Magnetic moment
$B=$ Magnetic field of intensity
$\theta =$ Angle between the magnetic field and magnetic axes of the bar magnet
Complete step by step solution:
A bar magnet of magnetic moment $M = 200A - {m^2}$ is suspended in a magnetic field of intensity, $B = 0.25N/A - m$ (given)
Now we know that the formula for the torque acting on a magnet in a uniform magnetic field is given as:
$\tau = \overrightarrow M \times \overrightarrow B = MB\sin \theta $ … (1)
This magnet experiences a couple (or torque) in the magnetic field. Now to find out the couple required to deflect it through $\theta = {30^ \circ }$ we will take help from equation (1).
$ \Rightarrow \tau = MB\sin \theta $
Substituting all the required values from the question in the above expression, we get
$ \Rightarrow \tau = (200) \times 0.25 \times \sin \left( {{{30}^ \circ }} \right)$
$ \Rightarrow \tau = 50 \times 0.5 = 25N - m$ $\left( {\therefore \sin {{30}^ \circ } = \dfrac{1}{2} = 0.5} \right)$
Thus, the couple required to deflect a bar magnet through ${30^ \circ }$ is $25N - m$.
Hence, the correct option is (B) $25N - m$ .
Therefore, the correct option is B.
Note:
The alpha particles, beta particles and protons are charged particles, thereby being deflected by magnetic fields. Only gamma rays which are actually neutron particles are not deflected by magnetic fields as these particles have no charge. Neutron does not contain any electrical charge particles.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

JEE Advanced 2024 Syllabus Weightage

Current Loop as Magnetic Dipole and Its Derivation for JEE
