moving-charges-and-magnetism Question 34

Question: Q. 10. The figure shows three infinitely long straight parallel current carrying conductors. Find the

(i) magnitude and direction of the net magnetic field at point A lying on conductor 1 ,

(ii) magnetic force on conductor 2.

U] [Foreign I, II, III, 2017]

Show Answer

Solution:

Ans. (i) Magnitude of magnetic field at A

1

Direction of magnetic field at A

(ii) Magnitude of magnetic force on conductor 21

Direction of magnitude force on conductor 21/2

(i) B2=μ04π2(3I)r=μ04π(6Ir) into the plane of the paper /()

B3=μ04π(2(4I)3r)=μ04π(8I3r) out of the plane of the paper/ ()

BA=B2B3 into the paper

=μ04π(1013r) into the paper /()

(ii) F21=μ04π2I(3I)r away from wire 1 (towards 3 )

F23=μ04π2×(3I)×(4I)2r away from 3 (towards 1 )

Fnet =F23F21 towards wire 1

=μ04π6(I)2r towards wire 1

3

[CBSE Marking Scheme 2017]

[I] Q. 11. Two long straight parallel conductors carry steady current I1 and I2 separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere. A [Delhi I, II, III 2015]

Ans. Fab= Force experienced by wire ’ a ’ of length ’ l ’ due to magnetic field of wire ’ b ‘.

Fba= Force experienced by wire ’ b ’ of length ’ l ’ due to magnetic field of wire ’ a ‘.

Ba= Magnetic field due to wire ’ a ‘.

Bb= Magnetic field due to wire ’ b ‘.

since

Ba=μ0I12πd F=i(l×B) Fba=I1lμ0I22πd

Similarly,

Fab=I1lμ0I22πd

The direction of force experienced by the wire ’ a ’ is toward the wire ’ b ‘. (As shown in the diagram). Similarly the direction of force experienced by the wire ’ b ’ is toward the wire ’ a ‘. Thus, the force is attractive.

If two long wires are placed in vacuum at a separation of 1 m, one Ampere would be defined as the current in each wire that would produce a force of 2×107Nm1 per unit length of wire.



Table of Contents

NCERT Chapter Video Solution

Dual Pane