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The 10-turn loop of wire shown in FIGURE P29.71 lies in a horizontal plane, parallel to a uniform horizontal magnetic field, and carries a 2.0 A current. The loop is free to rotate about a nonmagnetic axle through the center. A 50 g mass hangs from one edge of the loop. What magnetic field strength will prevent the loop from rotating about the axle?

Short Answer

Expert verified

The required magnetic field is0.12T.

Step by step solution

01

Step 1. Given information

The diagram is as follows:

The loop of wire has 10turns, the mass of the block is 50gand the loop carries a current of2.0A.

02

Step 2. Calculation of the torque on the loop due to the block

If the loop is not allowed to rotate about the axle, the net torque created on the loop due to both the magnetic field and the block has to be equal with each other.

The formula to calculate the torque about the axle on the loop due to the block is given by

1=mgl...........................(1)

Here, 1is the torque due to the block, mis the mass of the block, gis the acceleration due to gravity and lis the distance of the block from the axle.

03

Step 3. Calculation of the torque due to the magnetic field

There is no torque on the arms of the loop which are parallel to the magnetic field. Only the arms which are at right angles to the direction of the magnetic field will experience the torque due to the magnetic field.

The formula to calculate the torque on the loop due to the magnetic field is given by

2=2nidBl........................(2)

Here, 2is the torque due to the magnetic field, nis the number of turns in the loop, iis the current through the loop,dis the length of each arm perpendicular to the magnetic field andBis the magnetic field.

04

Step 4. Calculation of the magnetic field

Equate equation (1) and equation (2) and simplify to obtain the magnetic field.

mgl=2nidBlB=mg2nid................................(3)

Substitute 50gfor m, 9.80m/s2for g, 10for n,2.0Afor iand 10.0cmfor dinto equation (3) to calculate the required magnetic field.

B=50g1kg1000g9.80m/s22102.0A10.0cm1m100cm0.12T

05

Step 5. Final answer

The required magnetic field is0.12T.

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