A long wire carries a 3.2 amp current to the right. A 0.35 meter portion of the

Question

A long wire carries a 3.2 amp current to the right. A 0.35 meter portion of the wire lies within a region of magnetic field pointing into the page.

a. If the force on this portion of the wire is 0.25 N, how strong is the magnetic field?

b. In which direction is that 0.25 N force pulling the wire?

c. If a second wire goes through the magnetic field, parallel to the first wire and below the first wire, with a 2 amp current going in the opposite direction as the first current, what will the new force on the first wire (with the 3.2 amp current) be and in which direction? Let the distance between the wires be 10 cm.

Explanation / Answer

Part a) F=0.25 N, I=3.2 Amp, L=0.35.

We know the formula F=I LB sin 90 (please note that here magnetic field is perpendicular to length carrying current I)

therfore F=ILB

B=F/I*L

B=0.25/(3.2 *0.35)

B=0.25/1.12

B=0.2232 Tesla

Part b) direction of force is given by the left hand rule where the first vector is l and second vector is B.

so the direction of force is perpendicular to both of them and in upward direction.

If B is inward,I is towards right so F is upwards

Part C) Force on the first wire due to another wire parrallel to it is given by

F/L=((0/4)* I1* I2)/r²

where 0/4=10-7

I1=I2=3.2 Amp, r=10 cm=0.1 m

F/L=10-7 *3.2 * 3.2 /(0.1)2

F/L=1024 * 10-7 N/m

F=1024 *0.35 *10-7 N

F=358.4  *10-7 N

Since the current in the first is opposite to the current in second wire so the both wire will repel each other.So force would be in upward direction.

GOOD LUCK!!!

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