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Question

A square loop of uniform conducting wire is as shown in fig. A current I (in amperes) enters the loop from one end and exits the loop from opposite end as shown. The length of one side of the square loop is 1 meter. The wire has uniform cross-section area and uniform linear mass density. In four situations of column I, the loop is subjected to four different magnetic fields. Under the conditions of column I, match the column I with corresponding results of column II (B0 in column I is a positive non-zero constant)

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Solution

Case A:
B=B0^i
Force on the wire along x-axis or parallel to x-axis is zero.
Force is on wire along y-axis or parallel to y-axis.

F=i2lB(^j×^i)=i2lB0^k

Since there are two sides along x-axis or parallel to x-axis, net force is

F=2×(i2lB0^k)=ilB0^k

Also, since since both the forces are opposite and pass through the center of the loop, net torque on the loop is zero.

Case B:
B=B0^j
Force on the wire along y-axis or parallel to y-axis is zero.
Force is on wire along x-axis or parallel to x-axis.

F=i2lB(^i×^j)=i2lB0^k

Since there are two sides along y-axis or parallel to y-axis, net force is

F=2×(i2lB0^j)=ilB0^k

Also, since since both the forces are opposite and pass through the center of the loop, net torque on the loop is zero.

Case C:
It is a combination of Case A and Case B

Hence, net force and net torque on the loop is zero.

Case D:

B=B0^j

Force on two sides along x-axis or parallel to x-axis is

F1=ilB0(^i×^k)=ilB0^j

Force on two sides along y-axis or parallel to y-axis is

F2=ilB0(^j×^k)=ilB0^j

Resultant of F1 and F2 is F=F1+F2=2ilB012(^i+^j)

Torque due to F is zero since the net force passes through the center of the loop.

A3,4
V3,4
C2,3
D1,3

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