A rectangular coil of length $$0.12$$ m and width $$0.1m$$ having $$50$$ turns of wire is suspended vertically in a uniform magnetic field of strength $$0.2$$ $$Weber/m2$$ . The coil carries a current of $$2$$ A. If the plane of the coil is inclined at an angle of $$30$$° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

Question

A rectangular coil of length $$0.12$$ m and width $$0.1m$$ having $$50$$ turns of wire is suspended vertically in a uniform magnetic field of strength $$0.2$$ $$Weber/m2$$ .  The coil carries a current of $$2$$  A. If the plane of the coil is inclined at an angle of $$30$$° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

Infinity Learn · Accepted Answer

The required torque is τ $$=$$ NIAB $$sin$$ θ where N is the number of turns in the coil, I is the current through the coil, B is the uniform magnetic field, A is the area of the coil and θ is the angle between the direction of the magnetic field and normal to the plane of the coil. Here, N $$=$$ $$50$$, I $$=$$ $$2$$ A, A $$=$$ $$0.12$$ m ×  $$0.1$$ $$m=$$ $$0.012$$ $$m2B$$ $$=$$ $$0.2$$ Wb $$/m2$$ and  θ $$=$$ $$90$$° $$-$$ $$30$$° $$=$$ $$60$$° ∴  τ$$=(50)(2$$ $$A)0.012$$ $$m20.2$$ $$Wb/m2sin60$$°$$=0.20$$ Nm

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