Discussion
A rod of length 1 m rotates at 60 rad/s about one end in a field of 0.2 T parallel to the axis of rotation. The emf between its ends is
Answer: B
The emf is \( \tfrac{1}{2}B\omega l^2 \), that is \( 0.5 \times 0.2 \times 60 \times 1 = 6 \) V. The factor of one half arises because the velocity of a point on the rod grows linearly from zero at the axis, so its average is half the tip value.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development