Question

4. Two long, parallel wires carry currents of 11 = 3.18 A and 12 = 4.50 A in the direction indicated in the figure below. (Choose the line running from wire 1 to wire 2 as the positive x-direction and the line running upward as the positive y direction.) Find the magnitude and direction of the magnetic field at point P, located d = 20.0 cm above the wire carrying the 4.50-A current. Р 1 12

Answer 1

Solution :

Here we have :

I₁ = 3.18 A
I₂ = 4.50 A

d = 20 cm = 0.20 m

.

B1 Bnet B2 5 r 01 d 41 12

.

Here, δ = tan^-1[d/d] = 45^o

Therefore, θ = 90^o - δ = 90^o - 45^o = 45^o

.

And, r² = d² + d² = 2 d²

∴ r = √2 d = √2 (0.20 m) = 0.2828 m

.

Now, Magnetic field due to a current carrying conductor is given by :

$B=\frac{\mu _0I}{2\pi d}$

Therefore :

Hol1 B1 (47 x 10-)(3.18) 27(0.2828) = 2.25 x 10-6 T = 2.25 4T 21

And,

$B_2=\frac{\mu _0I_2}{2\pi d}=\frac{(4\pi \times 10^{-7})(4.50)}{2\pi (0.20)}=4.5\times 10^{-6}\ T=4.5\ \mu T$

.

Therefore, Magnitude of the net magnetic field will be given by :

According to the parallelogram law of vector addition.

$B_{net}=\sqrt{(B_1)^2+(B_2)^2+2(B_1)(B_2)cos\theta }$

$\therefore \ B_{net}=\sqrt{(2.25)^2+(4.5)^2+2(2.25)(4.5)cos(45) }=6.295\ \mu T=6.3\ \mu T$

.

Therefore, The magnitude of the net magnetic field at point P will be given as :

B_net = 6.3 μT = 6.3 x 10^-6 T

.

And, The angle made by the resultant vector with B₂ will be given by :

$\beta =\arctan \left [ \frac{B_1cos\theta }{B_2+B_1sin\theta } \right ]=\arctan \left [ \frac{(2.25)cos(45) }{(4.5)+(2.25)sin(45) } \right ]=14.64^o$

Therefore, The direction of net magnetic field will be : 14.64^o - Clockwise from negative x-axis or 165.36^o - Counterclockwise from positive x-axis.