A solenoid having 170 turns and a cross-sectional area of 6.72 cm2 carries a cur

Question

A solenoid having 170 turns and a cross-sectional area of 6.72 cm2 carries a current of 1.10 A .

Part A

If it is placed in a uniform 1.18 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented perpendicular to the field.

1 = ____ Nm

Part B

If it is placed in a uniform 1.18 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented parallel to the field.

2 = ____ Nm

Part C

If it is placed in a uniform 1.18 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented at 45.0 o with the field.

3 = ____ Nm

My Past WRONG answers where

A) 1 = n*i*A*B*cos(0) = 0 N-m

B) 2= n*i*A*B = 170*1.1*6.72*10^-4*1.18 = 0.148 N-m

C) 3 = 0.148*cos(45) = 0.104 N-m

Explanation / Answer

Here ,

magnetic moment of solenoid ,

u = N*I*A

u = 170 * 6.72 *10^-4 * 1.10

u = 0.126 A.m^2

as torque is given as

Torque = u * B * sin(theta)

A)

Now, for theta = 0 degree

T1 = u * B * sin(theta)

T1 = 0.126 * 1.18 * sin(0)

T1 = 0 N.m

the torque acting on the solenoid is 0 N.m

B)

now, for field parallel to solenoid ,

T2 = 0.126 * 1.18 * sin(90)

T2 = 0.148 N.m

the torque acting on the solenoid is 0.148 N.m

C)

now , for theta = 45 degree

T3 = 0.126 * 1.18 * sin(45)

T3 = 0.104 N.m

the torque acting on the solenoid is 0.104 N.m

these answers are correct

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