In Figure 29-46 two concentric circular loops of wire carrying current in the same direction lie in the same plane. Loop 1 has radius1.50cm and carries 4.00mA. Loop 2 has radius2.50cmand carries 6.00mA.Loop 2 is to be rotated about a diameter while the net magnetic field Bset up by the two loops at their common center is measured. Through what angle must loop 2 be rotated so that the magnitude of that net field is 100nT?

Short Answer

Expert verified

The angle through which loop 2 must be rotated so that the magnitude of the net field 100nTis144°.

Step by step solution

01

Given

  1. Radius of loop 1r1=1.50cm=1.50×10-2m.
  2. Current in the loop 1i1=4.00mA=4.00×10-3A.
  3. Radius of loop 2r2=2.50cm=2.50×10-2m.
  4. Current in the loop 2i2=6.00mA=6.00×10-3A.
  5. Net magnetic fieldB=100nT.
02

Understanding the concept

Use the formula of the magnetic field at the center of the coil due to two wires. From this find the angle through which loop 2 must be rotated so that the magnitude of the net filed is 100nT.

Formula:

B=μ0I2r
03

Calculate the angle through which loop 2 must be rotated so that the magnitude of the net field is 

For the magnetic field at the center of loop 1.

B1=μ0i12r1

B1=1.26×10-6T.mA4.00×10-3A21.50×10-2m

B1=1.68×10-7T=168nT

Now, for the magnetic field at the center of loop 2.

B2=μ0i22r2

Substitute the values and solve as:

B2=1.26×10-6T·mA6.00×10-3A22.50×10-2m

B2=1.512×10-7TB2=151.2nT

Now, we use the cosine rule of vector addition

B2=B12+B22+2B1B2cosθ

100nT2=168nT2+151.22+2168nT151.2cosθ

10000nT-51085.44nT=50803.2nTcosθ

50803.2nTcosθ=-41085.44nT

Solve further as:

cosθ=-41085.44nT50803.2nT

cosθ=-0.81

θ=cos-1-0.81

role="math" localid="1663002310798" θ=144°

Therefore, the angle through which loop 2 must be rotated so that the magnitude of the net field100nT is 144°.

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