Given:
- i2=200 μA i_2 = 200 \, \mu A i2=200μA,- θ2=60∘=π3 radians \theta_2 = 60^\circ = \frac{\pi}{3} \, \text{radians} θ2=60∘=3πradians.
The deflection θ \theta θ is proportional to the current i i i. Therefore:
i1i2=θ1θ2.\frac{i_1}{i_2} = \frac{\theta_1}{\theta_2}.i2i1=θ2θ1.
For θ1=π10 radians \theta_1 = \frac{\pi}{10} \, \text{radians} θ1=10πradians:
i1200=π10π3.\frac{i_1}{200} = \frac{\frac{\pi}{10}}{\frac{\pi}{3}}.200i1=3π10π.
Simplify:
i1200=310 ⟹ i1=200×310=60 μA.\frac{i_1}{200} = \frac{3}{10} \implies i_1 = 200 \times \frac{3}{10} = 60 \, \mu A.200i1=103⟹i1=200×103=60μA.The Correct answer is: 60 μA\mu AμA