Question

Long wire 1 carries 9.0 Amperes Westward on the table along the (-) x-axis; long wire 2 carries 3.0 Amperes Northward on the table along the (+) y-axis. At which of these locations is the magnetic field from these currents equal to zero? x = 30 mm, y = -90 mm x = 90 mm , y = 30 mm D x = 30 mm , y = 90 mm x = -90 mm , y = -10 mm x = -30 mm , y = 90 mm x = -90 mm , y = -30 mm x = -30 mmy = -90 mm x = 90 mm , y = -30 mm x = -10 mm y = 90 mm x = -90 mm y = 30 mm x = 10 mm , y = 90 mm x = 10 mm y = -90 mm

Answer 1

Magnetic field due to a long wire is given as

$B = \frac{\mu_0 I}{2\pi r}$

Due to wire 1

$B_1 = \frac{\mu_0 I_1}{2\pi y}$

Due to wire 2

$B_2 = \frac{\mu_0 I_2}{2\pi x}$

In the second and fourth quadrant B₁ and B₂ are in the opposite direction so we have net filed zero.

so in 2nd and 4th quadrant

$0= \frac{\mu_0 I_1}{2\pi y} - \frac{\mu_0 I_2}{2\pi x}$

$\frac{ I_2}{x} = \frac{ I_1}{y}$

$\frac{ 3}{x} = \frac{ 9}{y}$

So field is zero at (-30,90)mm

and at                   (30,-90)mm

ques 6

Since the magnetic field is uniform at the place of the loop so net force is zero

F = 0

torque is goven as

$\tau = NIAB sin(\theta)$

angle betweeb B and area vector of the loop = 90 - 20 = 70 so,

$\tau = (84)(0.5)(\pi \times (0.0286^2)) (50 \times 10^{-6})sin(70)$

$\tau =5.071 \mu T$