Question

A single-turn square loop carries a current of 16 A . The loop is 15 cm on a side and has a mass of 3.8×10^−2kg . Initially the loop lies flat on a horizontal tabletop. When a horizontal magnetic field is turned on, it is found that only one side of the loop experiences an upward force. Find the minimum magnetic field, Bmin, necessary to start tipping the loop up from the table. Express your answer using two significant figures.

Answer 1

Answer:

The minimum magnetic field is 0.078 T.

Explanation:

Given that,

Current = 16 A

Side = 15 cm

Mass $m= 3.8\times10^{-2}\ kg$

Mass each segment in given square loop is

$m=\dfrac{3.8\times10^{-2}}{4}$

We need to calculate the torque due to gravity

Using formula of torque

$\tau_{g}=2mg(\dfrac{L}{2})+mgL$

$\tau_{g}=2mgL$

The torque due to magnetic field

$\tau_{B}=FL$

$\tau_{B}=BIL^2$

The equilibrium condition

$\tau_{B}=\tau_{g}$

Put the value into the formula

$BIL^2=2mgL$

$B=\dfrac{2mgL}{IL^2}$

$B=\dfrac{2mg}{IL}$

Put the value into the formula

$B=\dfrac{2\times\dfrac{3.8\times10^{-2}}{4}\times9.8}{16\times15\times10^{-2}}$

$B=0.078\ T$

Hence, The minimum magnetic field is 0.078 T.