A conducting rod A C of length 4l is rotated about a point O in a uniform magnetic field vecB directed into the paper. A O=I and O C= 3l. Then

Question

A conducting rod A C of length 4l is rotated about a point O in a uniform magnetic field vecB directed into the paper. A O=I and O C= 3l. Then (A) VA-VO=(B ω l2/2) (B) VO-VC=(7/2) B ω l2 (C) VA-VC=4 B ω l2 (D) VC-VO=(9/2) B ω l2

Tardigrade Admin · Accepted Answer

By using e=(1/2) Bl2 ω For part A O ; e OA = e 0- e A =(1/2) B l2 ω For part O C ; e OC = e O - e C =(1/2) B (3 l )2 ω ∴ e A - e C = 4 B l2 ω

Q. A conducting rod $A C$ of length $4 l$ is rotated about a point $O$ in a uniform magnetic field $B$ directed into the paper. $A O = I$ and $OC = 3 l$ . Then

$V_{A} - V_{O} = \frac{B ω l ^{2}}{2}$

$V_{O} - V_{C} = \frac{7}{2} B ω l^{2}$

$V_{A} - V_{C} = 4 B ω l^{2}$

$V_{C} - V_{O} = \frac{9}{2} B ω l^{2}$

By using $e = \frac{1}{2} B l^{2} ω$
For part $A O; e_{O A} = e_{0} - e_{A} = \frac{1}{2} B l^{2} ω$
For part $OC; e_{OC} = e_{O} - e_{C} = \frac{1}{2} B (3 l)^{2} ω$
$∴ e_{A} - e_{C} = 4 B l^{2} ω$

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

VA​−VO​=2Bωl2​

VO​−VC​=27​Bωl2

VA​−VC​=4Bωl2

VC​−VO​=29​Bωl2