(C) The torque $\tau$ on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin(\theta)$,where $\theta$ is the angle between the

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The torque $\tau$ on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin(\theta)$,where $\theta$ is the angle between the normal to the coil and the magnetic field.Given: $N = 100$,$I = 1\,A$,$B = 0.5\,T$,and area $A = 20\,cm \times 20\,cm = 0.2\,m \times 0.2\,m = 0.04\,m^2$.Since the magnetic field is parallel to the plane of the coil,the angle between the normal to the coil and the magnetic field is $\theta = 90^\circ$.Therefore,$\tau = N I A B \sin(90^\circ) = 100 \times 1 \times 0.04 \times 0.5 \times 1$.$\tau = 100 \times 0.02 = 2\,N-m$.

Class 12 Physics Moving Charges and Magnetism Torque , Potential Energy and Work Done in Mangetic Field

Easy

$A$ rectangular coil $20\,cm \times 20\,cm$ has $100$ turns and carries a current of $1\,A$. It is placed in a uniform magnetic field $B = 0.5\,T$ with the direction of the magnetic field parallel to the plane of the coil. The magnitude of the torque required to hold this coil in this position is ........ $N-m$.

A
$0$
B
$200$
C
$2$
D
$10$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Moving Charges and Magnetism

Subtopic

Torque , Potential Energy and Work Done in Mangetic Field

$A$ thin magnetic needle is placed in a magnetic field of $200 \ G$ with its axis at $30^{\circ}$ to the direction of the field. Find the magnetic moment of the needle if it experiences a torque of $0.012 \ Nm$ in this field. (in $Am^2$)

EasyTS EAMCET 2022

View Solution

$A$ uniform magnetic field of $0.3 \text{ T}$ is established along the $+z$-direction. $A$ rectangular loop of sides $10 \text{ cm}$ and $5 \text{ cm}$ carries a current of $12 \text{ A}$ and is placed in the $xy$-plane as shown in the figure. The torque acting on the loop is $....... \text{ Nm}$.

EasyGSEB 2023

View Solution

$A$ current-carrying loop is free to turn in a uniform magnetic field. The loop will come into equilibrium when its plane is inclined at

Easy

View Solution

$A$ circular loop of area $0.01\,m^2$ carrying a current of $10\,A$ is held perpendicular to a magnetic field of intensity $0.1\,T$. The torque acting on the loop is......$N-m$.

EasyAIPMT 1994

View Solution

$A$ small coil of $N$ turns has an effective area $A$ and carries a current $I$. It is suspended in a horizontal magnetic field $\overrightarrow{B}$ such that its plane is perpendicular to $\overrightarrow{B}$. The work done in rotating it by $180^\circ$ about the vertical axis is

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ thin magnetic needle is placed in a magnetic field of $200 \ G$ with its axis at $30^{\circ}$ to the direction of the field. Find the magnetic moment of the needle if it experiences a torque of $0.012 \ Nm$ in this field. (in $Am^2$)

$A$ uniform magnetic field of $0.3 \text{ T}$ is established along the $+z$-direction. $A$ rectangular loop of sides $10 \text{ cm}$ and $5 \text{ cm}$ carries a current of $12 \text{ A}$ and is placed in the $xy$-plane as shown in the figure. The torque acting on the loop is $....... \text{ Nm}$.

$A$ current-carrying loop is free to turn in a uniform magnetic field. The loop will come into equilibrium when its plane is inclined at

$A$ circular loop of area $0.01\,m^2$ carrying a current of $10\,A$ is held perpendicular to a magnetic field of intensity $0.1\,T$. The torque acting on the loop is......$N-m$.

$A$ small coil of $N$ turns has an effective area $A$ and carries a current $I$. It is suspended in a horizontal magnetic field $\overrightarrow{B}$ such that its plane is perpendicular to $\overrightarrow{B}$. The work done in rotating it by $180^\circ$ about the vertical axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module