Question

A horizontal wire of length 0.11 m, carrying a current of 8.4 A, is placed in a uniform external magnetic field. When the wire is horizontal, it experiences no magnetic force. When the wire is tilted upward at an angle of 16 degree, it experiences a magnetic force of 2.5 times 10'3 N. Determine the magnitude of the external magnetic field. the tolerance is +/-2%

Answer 1

Answer 2

F = I*L*B*sin(theta)

2.5*10^-3 = 8.4*0.11*B*sin(16)

B = 9.816*10^-3N/mA

Answer 3

The wire experiences no magnetic force when kept horizontally because the magnetic field is horizontal and parallel to the wire.

F=I(LxB)=0.

When the wire is tilted, it makes angle $\theta$ =16⁰.

Magnitude of Force F = ILB sin $\theta$

B=F/(ILsin $\theta$ )

=2.5*10^-3/(8.4*0.11*sin 16) = 9.82*10^-3 T