
A coil $10cm$ diameter has $300turns$. If the coil carries a current of $10mA$ and lies in a magnetic field $5 \times {10^{ - 2}}T$, the maximum torque experienced by the coil is:
A) $4.7 \times {10^{ - 2}}Nm$
B) $4.7 \times {10^{ - 4}}Nm$
C) $4.7 \times {10^{ - 5}}Nm$
D) $4.7 \times {10^{ - 8}}Nm$
Answer
224.4k+ views
Hint: When a current flows inside the coil it generates a magnetic flux around the coal, now as the body is placed between a magnetic field, the magnetic field of the body and magnetic field in which it has been placed opposes each other and the body experiences a torque and tends to rotate.
Complete step by step solution:
The torque experienced by a current-carrying coil depends upon the current flowing through it, the number of turns, the area of the coil, and the magnetic field in which it is placed so,
$\therefore \tau = BIA\sin \theta \times n$
$B$ is the magnetic field in which the coil is placed
$I$ is the current flowing through the coil
$A$ is the cross-sectional area of the coil
$\theta $ is the angle at which the coil is placed
$n$ the number of turns in the coil
Here it is asked for the maximum torque and we know the maximum value on$\sin \theta = 1$at $\theta = 90^\circ $So the coil should make an angle of$90^\circ $with magnetic field
Now the various given values are
$B = 5 \times {10^{ - 2}}T$
$I = 10mA$ the value of current should be in ampere so we convert the value into amperes
$ \because 1A = 1000mA$
$\therefore 10mA = 0.01A$
$A = \pi {r^2}$
Here diameter is given as $2cm$ so the radius will be $1cm$but we need to convert this into the meter so, $ r = 0.01m$
$\therefore A = \pi \times {\left( {0.01} \right)^2}$
$\Rightarrow A = 3.14 \times {10^{ - 4}}$
Substituting all the values in the equation we get
\[
\tau = \left( {5 \times {{10}^{ - 2}}} \right) \times 0.01 \times \left( {3.14 \times {{10}^{ - 4}}} \right)\sin 90^\circ \times 300 \\
\Rightarrow \tau = 4.71 \times {10^{ - 5}}Nm \\
\]
Final answer is (C), the total torque experienced by the coil will be $4.71 \times {101^{ - 5}}Nm.$
Note: When the current flows through any conductor it generates a magnetic field around it. The magnetic field of the conductor when it comes in between other magnetic fields it experiences a torque which tends to rotate it, this is called the magnetic field of electric current. This is the explanation of what happens inside an electric motor.
The torque will increase by increasing the no of turns or the current passing through the coil.
Complete step by step solution:
The torque experienced by a current-carrying coil depends upon the current flowing through it, the number of turns, the area of the coil, and the magnetic field in which it is placed so,
$\therefore \tau = BIA\sin \theta \times n$
$B$ is the magnetic field in which the coil is placed
$I$ is the current flowing through the coil
$A$ is the cross-sectional area of the coil
$\theta $ is the angle at which the coil is placed
$n$ the number of turns in the coil
Here it is asked for the maximum torque and we know the maximum value on$\sin \theta = 1$at $\theta = 90^\circ $So the coil should make an angle of$90^\circ $with magnetic field
Now the various given values are
$B = 5 \times {10^{ - 2}}T$
$I = 10mA$ the value of current should be in ampere so we convert the value into amperes
$ \because 1A = 1000mA$
$\therefore 10mA = 0.01A$
$A = \pi {r^2}$
Here diameter is given as $2cm$ so the radius will be $1cm$but we need to convert this into the meter so, $ r = 0.01m$
$\therefore A = \pi \times {\left( {0.01} \right)^2}$
$\Rightarrow A = 3.14 \times {10^{ - 4}}$
Substituting all the values in the equation we get
\[
\tau = \left( {5 \times {{10}^{ - 2}}} \right) \times 0.01 \times \left( {3.14 \times {{10}^{ - 4}}} \right)\sin 90^\circ \times 300 \\
\Rightarrow \tau = 4.71 \times {10^{ - 5}}Nm \\
\]
Final answer is (C), the total torque experienced by the coil will be $4.71 \times {101^{ - 5}}Nm.$
Note: When the current flows through any conductor it generates a magnetic field around it. The magnetic field of the conductor when it comes in between other magnetic fields it experiences a torque which tends to rotate it, this is called the magnetic field of electric current. This is the explanation of what happens inside an electric motor.
The torque will increase by increasing the no of turns or the current passing through the coil.
Recently Updated Pages
JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: City Intimation Slip Releasing Today, Application Form Closed, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Hybridisation in Chemistry – Concept, Types & Applications

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Electromagnetic Waves and Their Importance

Understanding Atomic Structure for Beginners

