Question

A straight wire of length 50cm carrying a current of 2A is suspended freely in air in a uniform magnetic field of 0.5T. The mass of the wire is

Answer 1

Answer:

5.1 g

Explanation:

We are given that

Length  of straight wire=50 cm=$50\times 10^{-2} m$

Current flowing in wire=2 A

Magnetic field=0.5 T

We have to find the mass of the wire.

$F=IBL$

Substitute the values then we get

$F=2\times 0.5\times 5\times 10^{-2}$

$F=5\times 10^{-2} N$

We know that

$F=mg$

Where g=$9.8 m/s^2$

$5\times 10^{-2}=m\times 9.8$

$m=\frac{5\times 10^{-2}}{9.8}=0.51\times 10^{-2} Kg=5.1 g$

Hence, the mass of wire=5.1 g

Answer 2

The mass of the wire will be 0.051kg

Explanation:

The wire is pulled by gravitational force in the downward direction while the magnetic force will act in the upward direction. These two forces cancel each other. So, mass can be found by equating both forces.

mg = ilB

m = ilB/g

m = (2)(0.5)(0.5)/9.8

m =  0.051kg