Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A wooden stick of length 3l is rotated about an end with constant angular velocity $\omega$ in a uniform magnetic field B perpendicular to the plane of motion. If the upper one third of its length is coated with copper, the potential difference across the whole length of the stick is

Option: 1
$\frac{9\; B\; \omega l^{2}}{2}$

Option: 2
$\frac{4\; B\; \omega l^{2}}{2}$

Option: 3
$\frac{5\; B\; \omega l^{2}}{2}$

Option: 4
$\frac{\; B\; \omega l^{2}}{2}$

Answers (1)

best_answer

Motional E.m.f due to rotational motion -

Conducting rod $\rightarrow$

$\varepsilon =\frac{1}{2}Bl^{2}\omega =Bl^{2}\pi\nu$

$\nu \rightarrow f\! r\! equency$

$T \rightarrow Time\; period$

- wherein

When the rod rotates, there will be an induced current in the rod. The given situation can be treated as if a rod 'A' of length ' $3l$ ' rotating in the clockwise direction, while another say rod 'B' of length ' $2l$ ' rotating in the anti clockwise direction with same angular speed ' $\omega$ '.

As we know that $e=\frac{1}{2}B\omega l^{2}$

For 'A' :

$e_{A}=\frac{1}{2}B\omega \left ( 3l \right )^{2}$ & $e_{B}=\frac{1}{2}\; B\left ( -\omega \right )\left ( 2l \right )^{2}$

Resultant induced emf will be :

$e=e_{A}+e_{B}=\frac{1}{2}B\omega l^{2}\left ( 9-4 \right )$ $e=\frac{5}{2}B\; \omega l^{2}$

Posted by

Rishi

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Punjab’s new 2-year MBBS bond, fee hike ‘exploitation’, ‘unfair’; interns, doctors shut OPDs in protest

anam-khan

NEET UG 2025 counselling likely to begin on July 1; AIQ seat allocation, court hearing updates

anam-khan

NEET UG: Delhi HC seeks NTA's response on plea alleging defective biometric system at centre

Latest Question