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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 of30.0掳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

01

Definition of magnetic field

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

02

Determine the magnitude and direction of magnetic field

The loop in equilibrium so the net torque must be zero 鈭�蟿=0

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

Due to gravitational force

蟿mg=mgrsin鈦�(蠒)=mg(0.400m)sin鈦�30鈭�

Due to magnetic field

蟿B=IABm(60掳)

Net torque is zero so

IABsin60鈭�=mg(0.0400m)sin鈦�30鈭�

Here the value of m , A

m=(0.15g/cm)2(8.00cm+6.00cm)m=4.2gm=4.2脳10鈭�3kgAndA=(0.0800m)(0.0600m)A=4.80脳10鈭�3m2

Now put all values

B=mg(0.0400m)sin鈦�30鈭�IAsin鈦�60鈭�B=4.2脳10鈭�3kg9.80m/s2(0.0400m)sin鈦�30鈭�(8.2A)4.80脳10鈭�3m2sin鈦�60鈭�B=0.024T

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

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