Question

Two flat circular coils have a common center, but
their planes are at right angles to each other. The
inner coil has 150 turns and radius of ???? cm. The
outer coil has 400 turns and a radius of 2???? cm.
The magnitude of the resultant magnetic
induction at the common centers of the coils
when a current of 200 mA is sent through each of
them is
(a) 10⁻³ Wb/m² (b) 2 × 10⁻³ Wb/m²
(c) 5 × 10⁻³ Wb/m² (d) 7 × 10⁻³ Wb/m²

Answer 1

Answer:

Magnetic field at common center of the circular loop is

$B = 10^{-3} Wb/m^2$

Explanation:

Magnetic field at the center of the coil due to current flowing n it is given as

$B = \frac{\mu_0 N i}{2R}$

now for coil 1 we have

$R = \pi cm$

$N_1 = 150$

$i = 200 mA$

so we have

$B_1 = \frac{(4\pi \times 10^{-7})(150)(200\times 10^{-3})}{2\pi \times 10^{-2}}$

$B_1 = 0.6 \times 10^{-3} T$

Now for 2nd coil we have

$R = 2\pi cm$

$N_2 = 400$

$i = 200 mA$

so we have

$B_2 = \frac{(4\pi \times 10^{-7})(400)(200\times 10^{-3})}{2(2\pi) \times 10^{-2}}$

$B_2 = 0.8 \times 10^{-3} T$

Now net magnetic field at the center is given as

$B = \sqrt{B_1^2 + B_2^2}$

$B = \sqrt{0.6^2 + 0.8^2}\times 10^{-3}$

$B = 10^{-3} Wb/m^2$

#Learn

Topic: Magnetic field due to a circular loop

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