Question

A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is present, given by B = 4.0i + 9.0x2 j , with x in meters and B in mT. Calculate the k-component of the force on the 2 m segment of the conductor that lies between x = 1.0 m and x = 3.0 m.

Answer 1

0.546
`\hat k`

Step-by-step explanation:

From the given information:

The force on a given current-carrying conductor is:

`F = I ( \L \limits ^ {\to } * B ^(\to))\\ \\ dF = I(dL\limits ^ {\to } * B ^(\to))`

where the length usually in negative (x) direction can be computed as

`\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i`

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

`\int dF = \int ^3_1 I ( dL^(\to) * B ^(\to))`

`F = I \int^3_1 ( -dx \hat i ) * ( 4.0 \hat i + 9.0 \ x^2 \hat j)`

`F = I \int^3_1 - 9.0x^2 \ dx \hat k`

`F = I (9.0) \bigg [(x^3)/(3) \bigg ] ^3_1 \hat k`

`F = I (9.0) \bigg [(3^3)/(3) - (1^3)/(3) \bigg ] \hat k`

where;

current I = 7.0 A

`F = (7.0 \ A) (9.0) \bigg [(27)/(3) - (1)/(3) \bigg ] \hat k`

`F = (7.0 \ A) (9.0) \bigg [(26)/(3) \bigg ] \hat k`

F = 546 × 10⁻³ T/mT
`\hat k`

F = 0.546
`\hat k`