What is the force per meter of length on a straight wire carrying a 8.05 A current when perpendicular to a 0.82 T uniform magnetic field?

The required magnitude of force per meter of wire is of 6.601 N/m.

Given data:

The magnitude of current in a wire is, I = 8.05 A.

The strength of magnetic field is, B = 0.82 T.

We known that, in a magnetic field there is always a force acting on a wire kept in the region. This force is known as magnetic force, and the expression for the magnetic force is given as,

Now,

For the force per unit length, and current being perpendicular to magnetic field,

Thus, we can conclude that the required magnitude of force per meter of wire is of 6.601 N/m.

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6.601 N/m.

Explanation:

Generally, the force on a wire carrying current in a magnetic field can be determined by:

F = BIL Sinθ

where: F is the force, B is the value of the magnetic field, I is the value of current in the wire and θ is the angle between the direction of current and magnetic field.

For force per meter of length, we have;

= BI Sinθ

Given: I = 8.05 A, B = 0.82 T and θ = (since they are perpendicular).

Then;

= 0.82 x 8.05 x Sin

  = 6.601

  = 6.601 N/m

the force per meter length is 6.601 N/m.