JEE Physics 2024 Question with Solution

Given: charges are $-4\mu C$ and $+4\mu C$ placed at $A(1,0,4)$ and $B(2,-1,5)$ in an electric field $\vec{E}=0.20\hat{i}\,V/cm$.

Find: the value of $\alpha$ if the torque magnitude is $8\sqrt{\alpha}\times 10^{-5}\,Nm$.

The dipole moment vector is

$$\vec{p}=q\vec{d}$$

where $q=4\times 10^{-6}\,C$ and the displacement vector from $A$ to $B$ is

$$\vec{d}=(2-1)\hat{i}+(-1-0)\hat{j}+(5-4)\hat{k}=\hat{i}-\hat{j}+\hat{k}\,m$$

So,

$$\vec{p}=4\times 10^{-6}(\hat{i}-\hat{j}+\hat{k})\,Cm$$

Convert the electric field:

$$\vec{E}=0.20\hat{i}\,V/cm=20\hat{i}\,V/m$$

The torque on a dipole is

$$\vec{\tau}=\vec{p}\times \vec{E}$$

Thus,

$$\vec{\tau}=\left(4\times 10^{-6}(\hat{i}-\hat{j}+\hat{k})\right)\times 20\hat{i}$$ $$\vec{\tau}=8\times 10^{-5}\left[(\hat{i}\times \hat{i})-(\hat{j}\times \hat{i})+(\hat{k}\times \hat{i})\right]$$ $$\vec{\tau}=8\times 10^{-5}(\hat{j}+\hat{k})\,Nm$$

Therefore, its magnitude is

$$|\vec{\tau}|=8\times 10^{-5}\sqrt{1^2+1^2}=8\times 10^{-5}\sqrt{2}$$

Comparing with

$$|\vec{\tau}|=8\sqrt{\alpha}\times 10^{-5}\,Nm$$

we get

$$\alpha=2$$

Therefore, the required value is $2$.

Given: the dipole is formed by equal and opposite charges of magnitude $4\times 10^{-6}\,C$, and the electric field is along the $+x$-direction.

Find: the numerical value of $\alpha$.

From the coordinates,

$$\vec{AB}=\hat{i}-\hat{j}+\hat{k}$$

Hence the dipole moment has components along $\hat{i}$, $-\hat{j}$ and $\hat{k}$. Since the field is only along $\hat{i}$, only the components perpendicular to $\hat{i}$ contribute to torque.

So the effective perpendicular part is proportional to

$$-\hat{j}+\hat{k}$$

whose magnitude is

$$\sqrt{(-1)^2+1^2}=\sqrt{2}$$

Also,

$$|\vec{p}|_{\perp}=4\times 10^{-6}\sqrt{2}$$

and therefore

$$|\vec{\tau}|=|\vec{p}_{\perp}|E=(4\times 10^{-6}\sqrt{2})(20)=8\times 10^{-5}\sqrt{2}$$

Comparing with $8\sqrt{\alpha}\times 10^{-5}$ gives

$$\alpha=2$$

Therefore, the correct numerical answer is $2$.