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?

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;

[tex]\frac{F}{L}[/tex] = BI Sinθ

Given: I = 8.05 A, B = 0.82 T and θ = [tex]90^{o}[/tex](since they are perpendicular).

Then;

[tex]\frac{F}{L}[/tex] = 0.82 x 8.05 x Sin [tex]90^{o}[/tex]

= 6.601

= 6.601 N/m

the force per meter length is 6.601 N/m.

snehashish65 snehashish65 · Answer 2 · 2021-12-18T05:35:37+01:00

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,

[tex]F = B \times I \times L \times sin \theta[/tex]

Now,

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

[tex]\dfrac{F}{L} = B \times I \times sin90\\\\\dfrac{F}{L} = 0.82 \times 8.05 \times sin90\\\\\dfrac{F}{L} = 6.601 \;\rm N/m[/tex]

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

Learn more about the magnetic force here:

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