Two long wires carrying currents are 0.1 m apart as shown below in the figure. T

Question

Two long wires carrying currents are 0.1 m apart as shown below in the figure. The left wire has a current IL= 15 A (into the page), while the right wire has a current IR= 7 A (out of the page).

a. What is the magnitude of the magnetic field created by the left wire at the midpoint, C?

b.What is the direction of the magnetic field created by the left wire at the midpoint, C?

c.What is the magnitude of the magnetic field created by the right wire at the midpoint, C?

d.What is the direction of the magnetic field created by the right wire at the midpoint, C?

eWhat is the magnitude of the TOTAL magnetic field at the midpoint, C?

f.What is the direction of the TOTAL magnetic field at the midpoint, C?

gYou now place a 15 cm long wire at the midpoint, C, which has a current of 10 A OUT of the page. What is the magnitude of the force on this third wire?

Explanation / Answer

As the currents are in th opposite direction ,

the net magnetic field can be reduced.

B= mue0 *i /(2 *phi *r)

a) clock wise magnetic field i.e around the wire.

b) B1 = mue0 *15 /(2 *phi *0.05)

B1= 2 *10 ^-7*(15/0.05) = 600 *10^-7 T

c)anti-clock wise magnetic field around the wire.

d)B2 =mue0 *i /(2 *phi *r)

B2= 2*10 ^-7 * (7/0.05)= 140 *10 ^-7 T

e) the net magnetic field -s 460 *10 ^-7 T

f) clock wise direction around the wire.

g)Force at the mid point

F= mue0 *i1 i2 / (2 *phi * R)

due to first wire at left and due to second wire at right the force would exert on the middle wire.

F1 =mue0 *i1 i2 / (2 *phi * R)=2*10 ^-7 (i1 * i2 /R)

F1= 2*10 ^-7 (15 *10 )/0.05=3000* 10 ^-7 N

due to second wire which is at right

F2=mue0 *i1 i2 / (2 *phi * R)

= 2 *10 ^-7 * (7*10/0.05)=1400 X 10 ^-7 N

The net force at the middle is 1600 *10 ^-7 N

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