Question

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

Answer 1

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Step-by-step explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

`F=I * B * L * sin(\theta)`

So from data

`F = 5.20 A * 0.430 T * 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N`

Now sub parts

(a)

`\theta=60^(o)\\\\Force = 6.2608 sin(60^(o)) N\\\\Force = 5.42 N`

(b)

`\theta=90^(o)\\\\Force = 6.2608 sin(90^(o)) N\\\\Force = 6.26 N`

(c)

`\theta=120^(o)\\\\Force = 6.2608 sin(120^(o)) N\\\\Force = 5.42 N`