Question

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?

Answer 1

The current is 7.87 A.

Step-by-step explanation:

Given that,

Length = 0.250 m

Magnetic field
`B= 6.10*10^(-2)\ T`

Magnetic force = 0.120 N

We need to calculate the current

Using formula of magnetic force

`F=BIL`

`I=(F)/(BL)`

Where, B = magnetic field

I = current

l = length

F = force

Put the value into the formula

`I=(0.120)/(6.10*10^(-2)*0.250)`

`I=7.87\ A`

Hence, The current is 7.87 A.