A rectangular frame of wire abcd has dimensions 32cm X 8.0 cm and a total resistance of `2.0 Omega`. It is pulled out of a magnetic field `B = 0.020 T` by applying a force of `3.2 xx 10^(-5)` N . It is found that the frame moves with constant speed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between the points c and d.

Explanation

Here `R=2.0 Omega,`
`B=0.020 T,`
`l=32cm =0.32m,`
`b=8 cm =0.08 m`
`(a) F=i//B=3.2xx10^(-5)N`
` implies (B^2l^2v)/r=3.2xx10^(-5)`
` implies ((0.020)^2 xx(0.08)^2 xxv)/2=3.2xx10^(-5)`
` implies v= (32xx10^(-5))/(6.4xx10^(-3)xx4xx10^(-40)`
`=25 m//s.`
`(c) Resistance of part (ad)/(cd)=(2xx0.72)/(0.8)=1.8 Omega.`
`V_(ab)=iR=(Blv)/(2)xx1.8/2`
`=0.036V=3.6xx10^(-2)V`
`(d) Resistance of cd `
`=(2xx0.8)/0.8 =0.2 Omega`
`V=iR=(2xx0.08xx25xx0.2)/(2)`
`=4xx10^(-3)V.`