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a straight current-carrying wire in a uniform magnetic field, oriented at right angles to the wire. The 0.25-m wire carries a 4.75-A current and experiences a force of 0.38 N. What is the strength of the magnetic field?
0.45 T
0.56 T
0.32 T
0.72 T

Answer :

LEENAGO

Hello!

Using the equation for the magnetic force on a current-carrying wire:
[tex]F_B = \vec{B} \times i\vec{L}[/tex]

[tex]F_B[/tex] = Magnetic Force (0.38 N)
B = Magnetic field strength (T)
i = Current in wire (4.75 A)
L = Length of wire (0.25 m)

This is a cross product, so we can rewrite the equation as:
[tex]F_B = BiLsin\phi[/tex]

Where φ is the angle between the magnetic field vector and the length vector. In this instance, since the field is perpendicular to the wire, sin(90) = 1. We can simplify the equation to:
[tex]F_B = BiL[/tex]

Rearrange the equation to solve for 'B':
[tex]B = \frac{F_B}{iL}[/tex]

Plug in given values and solve.

[tex]B = \frac{0.38}{(4.75 * .25)} = \boxed{0.32 T}[/tex]

The strength of the magnetic field will be 0.32 T.Option B is correct.

What is magnetic field strength?

The number of magnetic flux lines on a unit area passing perpendicular to the given line direction is known as induced magnetic field strength .it is denoted by B.

Given data:

(I) current = 4.75-A

(B)induced magnetic field strrength=?

(L)length of the wire=0.25-m

(F)Magnetic force=0.38 N

The strength of the magnetic field is found as:

[tex]\rm F_B=BILsin \theta \\\\\ \theta = 90^0 \\\\ \rm F_B=BIL \\\\[/tex]

Substitute the values:

[tex]\rm 0.38 = B \times 4.75 \times 0.25 \\\\\ B= \frac{0.38}{4.75 \times 0.25 } \\\\\ B=0.32 \ T[/tex]

The strength of the magnetic field will be 0.32 T.

Hence, option B is correct.

To learn more about the strength of induced magnetic field refer ;

https://brainly.com/question/2248956

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