Question

4.A 31-cm long conducting wire of 9-g carrying 7- A current is placed in a uniform magnetic field. What are the strength and direction of the magnetic field needed to levitate the wire?

Answer 1

Answer

Given,

Length of the wire,L = 31 cm = 0.31 m

mass of the wire, m = 9 g = 0.009 Kg

Current in the wire,I = 7 A

Magnetic field strength, B= ?

Equating magnetic force to the weight of the wire.

$BIL = m g$

$B=\dfrac{m g}{IL}$

$B=\dfrac{0.009\times 9.81}{7\times 0.31}$

B = 0.0407 T

For Force to be upward magnetic field direction should be outward of the plane of paper.