Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Questions  >>  CBSE XII  >>  Physics  >>  Magnetism and matter
Answer
Comment
Share
Q)

A magnetic needle of magnetic moment $4.8 \times 10^{-2}\;JT^{-1}$ is placed at $30^{\circ}$ with the direction of uniform magnetic field of magnitude $3 \times 10^{-2} \;T$ Calculate the torque acting on the needle

$\begin{array}{1 1} 7.2 \times 10^{-4}J \\ 7.2 \times 10^{-4}JT^{-1} \\ 7.2 \times 10^{4}J \\ 7.2 \times 10^{4}JT^{-1} \end{array} $

1 Answer

Comment
A)
Solution :
Given : $B= 3 \times 10^{-2}\;T$
$\theta= 30^{\circ}$
$M= 4.8 \times 10^{-2}$
Torque $T= MB \sin \theta$
$\quad= 4.8 \times 10^{-2}$
$\quad= 4.8 \times 10^{-2} \times 3 \times 10^{-2} \times \sin 30^{\circ}$
$\quad= 4.8 \times 10^{-2} \times 3 \times 10^{-2} \times \large\frac{1}{2}$
$\quad= 7.2 \times 10^{-4}J$
Answer : $ 7.2 \times 10^{-4}J$
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...