Figure 17 shows a right-angled isosceles triangle PQR having its base equal to ‘a’. A current of…
Physics20223 marksShort answer
Figure 17 shows a right-angled isosceles triangle PQR having its base equal to ‘a’. A current of 1.0 A is passing downwards along a thin straight wire cutting the plane of a paper normally as shown at Q. Likewise, a similar wire carries an equal current moving normally upwards at R. Assume the wire is to be infinitely long.
(i)[1.0]
The magnitude and the direction of the magnetic induction B at P due to wire at ‘Q’:
(a)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ acting along PQ
(b)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ acting along PR
(c)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ towards the mid-point of QR
(d)$B = \frac{\mu_0 I}{\pi a}$ towards the mid-point of QR
(ii)[1.0]
The magnitude and the direction of the magnetic induction B at P due to wire at ‘R’:
(a)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ acting along PQ
(b)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ acting along PR
(c)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ towards the mid-point of QR
(d)$B = \frac{\mu_0 I}{\pi a}$ towards the mid-point of QR
(iii)[1.0]
The net magnitude and the direction of the magnetic induction B at P:
(a)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ acting along PQ
(b)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ acting along PR
(c)$B = \frac{\mu_0 I}{\sqrt{2} \pi a}$ towards the mid-point of QR
(d)$B = \frac{\mu_0 I}{\pi a}$ towards the mid-point of QR
