A conducting rod AC of length 4l is rotated about Ba→ point O in a uniform magnetic field B directed into the paper. AO=l and OC=3l then :

Question

A conducting rod AC of length 4l is rotated about Ba→ point O in a uniform magnetic field B directed into the paper. AO=l and OC=3l then : (A)  VA-Vo=(Bω l2/2)  (B)  Vo-Vc=(9/2)Bω l2  (C)  VA-Vo= β ω l22  (D)  Vc-Vo=(9/2) ω l2

Tardigrade Admin · Accepted Answer

VO-VA=(B ω l/2) × l =(B ω l2/2) VO-VC=(B ω 3 l/2) × 3 l =(9 B ω l2/2) Hence, VA-VC=4 B ω l2 or VA > VC

Q. A conducting rod AC of length $4 l$ is rotated about $_{B}^{a} \to$ point $O$ in a uniform magnetic field $B$ directed into the paper. $A O = l$ and $OC = 3 l$ then :

$V A - V o = \frac{B ω l ^{2}}{2}$

$V o - V c = \frac{9}{2} B ω l^{2}$

$V_{A} - V_{o} = β ω l^{2}^{2}$

$V_{c} - V_{o} = \frac{9}{2} ω l^{2}$

$V_{O} - V_{A} = \frac{B ω l}{2} \times l$
$= \frac{B ω l ^{2}}{2}$
$V_{O} - V_{C} = \frac{B ω 3 l}{2} \times 3 l$
$= \frac{9 B ω l ^{2}}{2}$
Hence, $V_{A} - V_{C} = 4 B ω l^{2}$
or $V_{A} > V_{C}$

Q. A conducting rod AC of length 4l is rotated about Ba​→ point O in a uniform magnetic field B directed into the paper. AO=l and OC=3l then :

VA−Vo=2Bωl2​

Vo−Vc=29​Bωl2

VA​−Vo​=βωl22

Vc​−Vo​=29​ωl2