Question

A wire 50 cm long with an east-west orientation carries a current of 7.0 A eastward. There is a uniform magnetic field perpendicular to this wire. If the force on the wire is 0.5 N upward, what are the direction and magnitude of the magnetic field?

Answer 1

The magnetic field is 0.143 T in north direction.

Step-by-step explanation:

Given that,

Length = 50 cm

Current = 7.0 A

Force = 0.5 N

We need to calculate the magnetic field

Using formula of magnetic force

`\vec{F}I\vec{L}*\vec{B}`

`F=ILB`

`B=(F)/(IL)`

`B=(0.5)/(7.0*50*10^(-2))`

`B=0.143\ T`

The direction of the magnetic field toward north.

Hence, The magnetic field is 0.143 T in north direction.

Answer 2

North, 0.1428 Tesla

Step-by-step explanation:

As we know that the magnetic force in the wire can be calculated as,

`F=Bilsin(\Theta )`.

Here the uniform magnetic field is perpendicular to wire than theta will be 90°.

So the magnetic force will become,

`F=Bil`.

Substitute 50 cm for l, 7 A for i, and 0.5 N for F in the above equation.

`B=(0.5)/(50* 1`0^(-2) m(7 A) )\\B=0.1428 T`

Now according to Fleming's left hand law, if the current in the east ward direction and force is in upward direction then the value of magnetic field will be perpendicular to both in the north direction.

Therefore magnetic field will be 0.1428 T in the north direction.