Question

QUESTION 1

Answer 1

Solution :

Magnetic field at the centre due to $I_P$ :

$B_1 = \frac{\mu_0 I_P}{2 \pi d}$

$B_1 = \frac{4 \pi \times 10^{-7} \times 0.2}{2 \pi d}$

$B_1 =10^{-7} \ T$

Its direction will be downwards in the plane of the paper.

Magnetic field at the centre due to $I_Q$ :

$B_2 = \frac{\mu_0 I_Q}{2 \pi d}$

$B_2 = \frac{4 \pi \times 10^{-7} \times 0.6}{2 \pi \times 0.8}$

$B_2 =1.5 \times 10^{-7} \ T$

Its direction will be upwards in the plane of the paper

Magnetic field due to the coil:

$B_3 = \frac{\mu_0 I_{coil}}{2 r}$

$B_3 = \frac{4 \pi \times 10^{-7} \times 0.5}{2 \times 0.2}$

$B_3 =3.14 \times 10^{-7} \ T$

Its direction will be rightwards.

Now the resultant of the magnetic field at the centre.

$B_{net} = \sqrt{(B_2 -B_1)^2+B^2_3}$

$B_{net} = \sqrt{(0.5 \times 10^{-7})^2+(3.14 \times 10^{-7})^2}$

$B_{net} = 3.18 \times 10^{-7} \ T$

Now the direction ,

$\tan \theta = \frac{B_2-B_1}{B_3}$

$\tan \theta = \frac{0.5 \times 10^{-7}}{3.14 \times 10^{-7}}$

Therefore $\theta = 9^\circ$ (from right to upward)