JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 3)
A rectangular coil (Dimension 5 cm × 2.5 cm)
with 100 turns, carrying a current of 3 A in the
clock-wise direction is kept centered at the
origin and in the X-Z plane. A magnetic field
of 1 T is applied along X-axis. If the coil is tilted
through 45° about Z-axis, then the torque on
the coil is :
0.42 Nm
0.55 Nm
0.38 Nm
0.27
Nm
Explanation
$$\left| {\overrightarrow \tau } \right| = \left| {\overline M \times \overline B } \right|$$
$$\tau = NI \times A \times B \times \sin {45^o}$$
$$\tau = 0.27 \,Nm$$
$$\tau = NI \times A \times B \times \sin {45^o}$$
$$\tau = 0.27 \,Nm$$
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