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A horizontal rod 0.250 m long is mounted on a balance and carries a current. At the location of the rod a uniform horizontal magnetic field has a magnitude of 6.10×10−2 T and direction perpendicular to the rod. The magnetic force on the rod is measured by the balance and is found to be 0.120 N . What is the current?

Respuesta :

Answer:

The current is 7.87 A.

Explanation:

Given that,

Length = 0.250 m

Magnetic field [tex]B= 6.10\times10^{-2}\ T[/tex]

Magnetic force = 0.120 N

We need to calculate the current

Using formula of magnetic force

[tex]F=BIL[/tex]

[tex]I=\dfrac{F}{BL}[/tex]

Where, B = magnetic field

I = current

l = length

F = force

Put the value into the formula

[tex]I=\dfrac{0.120}{6.10\times10^{-2}\times0.250}[/tex]

[tex]I=7.87\ A[/tex]

Hence, The current is 7.87 A.

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