A long straight wire is fixed horizontally and carries a current of $50 A$ . A second wire having linear mass density $1.0 \times 10^{- 4} kg / m$ is placed parallel to and directly above this wire at a separation of $5.0 mm$ . The second wire carries a current [[1]] $A$ in the opposite direction such that the magnetic repulsion can balance its weight.

Question

A long straight wire is fixed horizontally and carries a current of $50 A$ . A second wire having linear mass density $1.0 \times 10^{- 4} kg / m$ is placed parallel to and directly above this wire at a separation of $5.0 mm$ . The second wire carries a current [[1]] $A$ in the opposite direction such that the magnetic repulsion can balance its weight.

Accepted Answer

A long straight wire is fixed horizontally and carries a current of

50 A

. A second wire having linear mass density

1.0 \times 10^{- 4} kg / m

is placed parallel to and directly above this wire at a separation of

5.0 mm

. The second wire carries a current of

0.49 A

in the opposite direction such that the magnetic repulsion can balance its weight.
Given,
Current through one wire,

i_{1} = 50 A

Separation between the wires,

r = 5.0 mm

Converting it to meter, we know that

1 mm = 10^{- 3} m

So,

r = 5 mm = 5 \times 10^{- 3} m

Linear mass density of second wire =

1.0 \times 10^{- 4} kg / m

Magnetic field is the region in which electric charge experience magnetic influence. A straight wire carrying current will experience a magnetic field due to the current flowing through it.
The magnetic field due to two current-carrying wires is given by

B = \frac{μ_{0} i_{1} i_{2}}{2 πr}

In order that magnetic repulsion can balance the weight of the wire,

B = mg

⇨

\frac{μ_{0} i_{1} i_{2}}{2 πr} = mg

⇨

\frac{4 π \times 10^{- 7} \times 50 \times i_{2}}{2 π \times 5 \times 10^{- 3}} = 1.0 \times 10^{- 4} \times 9.8

⇨

2 \times 10^{- 3} \times i_{2} = 9.8 \times 10^{- 4}

⇨

i_{2} = \frac{9.8 \times 10^{- 4}}{2 \times 10^{- 3}}

⇨

i_{2} = 0.49 A

Hence, the current on the second wire is

0.49 A

.

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