Question

A straight conductor of mass m and carrying a current i is hinged at one end and placed in a plane perpendicular to the magnetic field of intensity as shown in the figure. At any moment if the conductor is let free, then the angular acceleration of the conductor will be : (Assume gravity free region).

• A. $\frac { 2 \mathrm { iB } } { 3 \mathrm {~m} }$
• B. $\frac { 3 \mathrm { iB } } { 2 \mathrm {~m} }$
• C.
• D. $\frac { 3 \mathrm { i } } { 3 \mathrm { mB } }$

Answer 1

Answer: (B)

$\frac { 3 \mathrm { iB } } { 2 \mathrm {~m} }$

Answer 2

The force acting on the elementary portion of the current carrying conductor is given as,

dF = i(d)B sin 90⁰

The torque applied by dF about O = dτ = rdF

⇒ The total torque about O = τ $\int \mathrm { d } \tau = \int \mathrm { r } ( \mathrm { iBdr } )$

$\Rightarrow \tau = \mathrm { iB } \int _ { 0 } ^ { \mathrm { L } } \mathrm { rdr } = \frac { \mathrm { iBL } ^ { 2 } } { 2 }$

The angular acceleration α = (Where M.I. = moment of inertia)

$\Rightarrow \alpha = \left( \frac { \mathrm { iBL } ^ { 2 } } { 2 } \right) / \left( \frac { \mathrm { mL } ^ { 2 } } { 3 } \right)$

_.

Hence (2) is correct.