91Ó°ÊÓ

We are given: - Radius of the disc, \(R=1 \mathrm{~m}\) - Charge on the disc, \(Q\) - Magnetic field \(B= k x t^{2}\) - Time \(t = 15 \mathrm{~s}\) - \(4kQ = 1\) S.I. unit

Torque, \(\tau\) on a segment of the disc can be defined as the cross product of the distance from the center and the magnetic force applied on that segment: \(d\tau = \vec{r} \times d\vec{F}\)

Using the force on a current loop formula, \(dF = Idl \times \vec{B}\), we can calculate the force on a small segment of the disc. We know that \(I = \frac{dq}{dt}\), where \(dq\) is the charge on the small segment of the disc and \(dt\) is the time. Let us consider a small annular section at a distance \(r\) from the center with a width \(dr\). The charge in this section will be \(dq = \frac{Q}{\pi R^2} \times 2 \pi r dr\). Thus, the force acting on this small annular section, \(dF = \frac{dq}{dt} \times dl \times B\). Here, \(dl = 2 \pi r\), the length of the annular section. Substituting the charge and length of the annular section into the force formula, \(dF = \frac{\frac{Q}{\pi R^2} \times 2 \pi r dr}{dt} \times 2 \pi r \times k x t^2\)

Now, we will calculate the torque on a small annular section using \(d\tau = \vec{r} \times d\vec{F}\), \(d\tau = r \times dF\) Substituting the value of \(dF\) from the previous step, \(d\tau = r \times \frac{\frac{Q}{\pi R^2} \times 2 \pi r dr}{dt} \times 2 \pi r \times k x t^2\)

To find the total torque acting on the disc, we will integrate the torque over the entire disc from \(r = 0\) to \(r = R\): \(\tau = \int_{0}^{R} r \times \frac{\frac{Q}{\pi R^2} \times 2 \pi r dr}{dt} \times 2 \pi r \times k x t^2\) As we are given the values of \(4kQ = 1 \mathrm{~S.I.}\) unit, \(R = 1 \mathrm{~m}\), and \(t = 15 \mathrm{~s}\), the integration can be simplified and calculated: \(\tau = \int_{0}^{1} r \times \frac{\frac{1}{4\pi} \times 2 \pi r dr}{15} \times 2 \pi r \times k x (15)^2\) \(\tau = \int_{0}^{1} r \times \frac{r dr}{30} \times 2 \pi r \times \frac{1}{4(15)^2}\) \(\tau = \frac{\pi}{90} \int_{0}^{1} r^3 dr\) \(\tau = \frac{\pi}{90} \left[ \frac{r^4}{4} \right]_{0}^{1}\) \(\tau = \frac{\pi}{90} \times \frac{1}{4}\) Finally, we get the torque as: \(\tau =\frac{\pi}{360}~N~m\) The torque acting on the disc after \(15\) seconds is \(\frac{\pi}{360}~N~m\).