Question

A straight, horizontal length of copper wire has a current i =50A through it.What are the magnitude and direction of the minimum magnetic field needed to suspend the wire when the length of wire is 14cm and mass is 40g.

Answer 1

Answer:

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Answer 2

Answer:

The magnetic field will act opposite to the weight of the wire i.e.mg. And the magnitude of the minimum magnetic field needed to suspend the wire is 5.71×10⁻³T.

Explanation:

We will use the following formulas to solve this question,

$F=BIL$      (1)

$F=mg$        (2)

Where,

F=force acting

B=magnetic field that is present

I=current flowing through the wire

L=length of the wire

m=mass of the wire

g=acceleration due to gravity=10m/s²

From the question we have,

I=50A

m=40g=40×10⁻³kg

L=14cm=14×10⁻²m

By equating equation (1) and (2) we get;

$BIL=mg$

$B=\frac{mg}{IL}$       (3)

By substituting the required values in equation (3) we get;

$B=\frac{40\times 10^-^3\times 10}{50\times 14\times 10^-^2}$

$B=\frac{40}{50\times 14}$

$B=\frac{4}{70}$

$B=0.0571=5.71\times 10^-^3T$

Since no external force is acting on the wire the magnetic field will act opposite to the weight of the wire i.e.mg. And the magnitude of the minimum magnetic field needed to suspend the wire is 5.71×10⁻³T.