Question

Magnetic field in the cylindrical region of radius R as shown in figure is given as $B=B_{0}t$. A charge q is taken from O to D via various path, then

• A. work done through path OACD by electric field is $\frac{2}{3}\pi B_{0}q{R}^2$.
• B. work done through path OABD by electric filed is negative.
• C. work done through straight path OCD by electric field is zero.
• D. work done by electric filed through path ACDB in close loop is zero.

Answer 1

Answer: (A), (C)

The correct options are
A work done through path OACD by electric field is $\frac{2}{3}\pi B_{0}q{R}^2$.
C work done through straight path OCD by electric field is zero.

$\int E.dl=\frac{d\varphi }{dt}=\frac{d}{dt}B.t\pi r^2$

$E=\frac{B_{0}r}{2}soF=qE=\frac{B_{0}qr}{2}$


Since field E is along tangent work done in path OA, AB, OC and CD is zero.

$W_{OACD}=W_{AC}=F\times d$

$=\frac{B_{0}q}{2}.2R.\frac{2\pi 2R}{6}=\frac{2}{3}\pi B_{0}q{R}^2$