Consider a flat, circular current loop of radius/carrying current I. Choose the

Question

Consider a flat, circular current loop of radius/carrying current I. Choose the x axis to be along the axis of the loop, with the origin at the center of the loop. Plot a graph of the ratio of the magnitude of the magnetic field at coordinate x to that at the origin, for x = 0 to x = 5 R. It may be useful to use a programmable calculator or a computer to solve this problem. Two very long, straight, parallel wires carry currents that are directed perpendicular to the page as shown in Figure P22.29. Wire 1 carries a current I_1 into the page (in the -z direction) and passes through the x axis at x = +a Wire 2 passes through the x axis at x = -2a and carries an unknown current I+2. the total magnetic field at the origin due to the current-carrying wires has the magnitude 2 mu_0 I_1/(2 pi a) the current I_2 can have either of two possible values, (a) Find the value of I_2 with the smaller magnitude, stating it in terms of I_2 and giving its direction, Find the other possible value of I_2. One very long wire carries current 30.0 A to the left along the x axis. A second very long wire carries current 50.0 A to the right along the line (y = 0.280 m, z = 0) Where in the plane of the two wires is the total magnetic field equal to zero? A particle with a charge of - 2.00 mu C is moving with a velocity of 150i Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle, A uniform electric field is applied to allow this particle to pass through this region unelected. Calculate the required vector electric field. A current path shaped as shown in Figure P22.31 produces a magnetic field at P, the center of the arc. If the are sub- tends an angle of 30.0degree and the radius of the area is 0.600 m

Explanation / Answer

Due to current I1 ,magnetic field at origin is B1 = mu_o*I1/(2*pi*a) along +Y-axis

Due to current I2,magnetic field at origin is B2 = mu_o*I2/(2*pi*2a),either + Y- or -Y-axis

Net magnetic field at the origin is Bnet = 2*mu_0*I1/(2*pi*a)

if I2 is out of the page then

B2 is along +y-axis

then Bnet = B1+B2 = (mu_o*I1/(2*pi*a)) + (mu_0*I2/(2*pi*2a)) = 2*mu_0*I1/(2*pi*a)

I1 + (I2/2) = 2*I1

I2/2 = I1

I2 = 2*I1

If I2 is into the page

B2 is along -Y-axis

Bnet = (mu_o*I1/(2*pi*a)) - (mu_0*I2/(2*pi*2a)) = 2*mu_0*I1/(2*pi*a)

I1 - (I2/2) = 2*I1

I2/2 = -I1

I2 = -2I1

answers for a) is I2 = -2*I1 direction is into the page

b) other possible value is I2 = 2*I1 direction is out of the page

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.