Chapter 4: Q69P (page 917)

The rectangular loop of wire shown in Figure has a mass of 0.15g per centimeter of length and is pivoted about side ab on a frictionless axis. The current in the wire is 8.2 A in the direction shown. Find the magnitude and direction of the magnetic field parallel to the -axis that will cause the loop to swing up until its plane makes an angle of $30.0 °$ with the yz -plane.

Short Answer

Expert verified

The magnitude of the magnetic field is 0.024 T and direction is y.

Step by step solution

Definition of magnetic field

The term magnetic field may be define as the area around the magnet behave like a magnet.

Determine the magnitude and direction of magnetic field

The loop in equilibrium so the net torque must be zero $\sum_{} τ = 0$

There is torque two type one is due to gravitation force and another is due to magnetic field.

Due to gravitational force

$\begin{aligned} τ_{mg} = mgrsin (ϕ) \\ = m g (0.400 m) \sin (30^{\circ}) \end{aligned}$

Due to magnetic field

$τ_{B} = IABm (60 °)$

Net torque is zero so

$IABsin (60^{\circ}) = m g (0.0400 m) \sin (30^{\circ})$

Here the value of m , A

$\begin{aligned} m = (0 .15 g / cm) 2 (8 .00 cm + 6 .00 cm) \\ m = 4 .2 g \\ m = 4 .2 \times 10^{- 3} kg \\ And \\ A = (0 .0800 m) (0 .0600 m) \\ A = 4 .80 \times 10^{- 3} m^{2} \end{aligned}$

Now put all values

$\begin{aligned} B = \frac{mg (0 .0400 m) (\sin (30^{\circ}))}{IAsin (60^{\circ})} \\ B = \frac{(4 .2 \times 10^{- 3} kg) (9 .80 m / s^{2}) (0 .0400 m) (\sin (30^{\circ}))}{(8 .2 A) (4 .80 \times 10^{- 3} m^{2}) \sin (60^{\circ})} \\ B = 0 .024 T \end{aligned}$