When a ‘J’ shaped conducting rod is rotating in its own plane with constant angular velocity ω about one of its ends P, in a uniform magnetic field \vec{B} (directed normally into the plane of paper) then magnitude of emf induced across it will be

 

  • Option 1)

    Bw\sqrt{L^{2}+l^{2}}

  • Option 2)

    \frac{1}{2}BwL^{2}

  • Option 3)

    \frac{1}{2}Bw\left ( L^{2}+l^{2} \right )

  • Option 4)

    \frac{1}{2}Bwl^{2}

 

Answers (1)

As learnt in

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

 

 This system can be replaced by a single rod of length = \sqrt{l^{2}+L^{2}}

\therefore  E.m.f induced = \frac{1}{2}B\omega \left ( l^{2}+L^{2} \right )


Option 1)

Bw\sqrt{L^{2}+l^{2}}

This option is incorrect

Option 2)

\frac{1}{2}BwL^{2}

This option is incorrect

Option 3)

\frac{1}{2}Bw\left ( L^{2}+l^{2} \right )

This option is correct

Option 4)

\frac{1}{2}Bwl^{2}

This option is incorrect

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions