Question

As shown in the figure, the rod is moving with velocity, $2\text{m/s}$. What will the value of velocity $v$ after time $t$ ? (Given: $B=3\text{T}$, and length of rod $=1\text{m}$, mass of rod $=m)$.

• A. $2e^{-3t/m}$
• B. $2e^{3t/m}$
• C. $2e^{6t/m}$
• D. $2-\frac{6}{m}t$

Answer 1

The correct option is D $2-\frac{6}{m}t$

The emf induced in a straight conductor of length $l$ moving with velocity $v$ in a magnetic field of intensity $B$ is given by,

$E=Blv$

The resistors are forming a Wheatstone Bridge.

$\therefore$ The equivalent resistance between $P$ and $Q$ is,

$R_{PQ}=\frac{6}{2}=3\Omega$

Current in the circuit,

$i=\frac{E}{R_{PQ}}$ (clockwise)

$\Rightarrow i=\frac{Blv}{R_{PQ}}$

$\therefore$ Force experienced by the moving conductor due to magnetic field,

${\overrightarrow{F}}_B=i(\overrightarrow{l}\times \overrightarrow{B})=ilB$

$\Rightarrow F_B=\frac{\left(Blv\right)\times \left(lB\right)}{R_{PQ}}=\frac{B^2{l}^{2}v}{R_{PQ}}$

Substituting the values,

$F_B=\frac{3^2\times 1^2\times 2}{3}=6\text{N}$ (Opposite to velocity )

Now, acceleration,

$a=\frac{-F_B}{m}$

$\Rightarrow {\int }_2^{v}dv=\frac{-6}{m}{\int }_0^{t}dt$

$\Rightarrow v-2=\frac{-6}{m}t$

$\therefore v=2-\frac{6}{m}t$

Hence, option $\left(D\right)$ is correct.