We are tasked with finding the net magnetic induction at the center of two concentric circular coils with the following properties:
1. Coil 1: Radius R1=20 cm=0.2 mR_1 = 20 \, \text{cm} = 0.2 \, \text{m}, current I1=0.2 AI_1 = 0.2 \, \text{A}, turns N1=10N_1 = 10.
2. Coil 2: Radius R2=40 cm=0.4 mR_2 = 40 \, \text{cm} = 0.4 \, \text{m}, current I2=0.4 AI_2 = 0.4 \, \text{A}, turns N2=10N_2 = 10.
3. The currents in the two coils flow in opposite directions, so the magnetic fields due to the coils will subtract.
Key Concept:
The magnetic field at the center of a circular coil carrying current II with NN turns and radius RR is given by:
B=μ0NI2RB = \frac{{\mu_0 N I}}{{2 R}}
Where μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}.
Step 1: Magnetic field due to Coil 1
For Coil 1:
B1=μ0N1I12R1B_1 = \frac{\mu_0 N_1 I_1}{2 R_1}
Substitute N1=10N_1 = 10, I1=0.2 AI_1 = 0.2 \, \text{A}, R1=0.2 mR_1 = 0.2 \, \text{m}:
B1=(4π×10−7)(10)(0.2)2(0.2)B_1 = \frac{(4\pi \times 10^{-7})(10)(0.2)}{2(0.2)} B1=(4π×10−7)(2)0.4B_1 = \frac{(4\pi \times 10^{-7})(2)}{0.4} B1=2π×10−6 TB_1 = 2\pi \times 10^{-6} \, \text{T}
Step 2: Magnetic field due to Coil 2
For Coil 2:
B2=μ0N2I22R2B_2 = \frac{\mu_0 N_2 I_2}{2 R_2}
Substitute N2=10N_2 = 10, I2=0.4 AI_2 = 0.4 \, \text{A}, R2=0.4 mR_2 = 0.4 \, \text{m}:
B2=(4π×10−7)(10)(0.4)2(0.4)B_2 = \frac{(4\pi \times 10^{-7})(10)(0.4)}{2(0.4)} B2=(4π×10−7)(4)0.8B_2 = \frac{(4\pi \times 10^{-7})(4)}{0.8} B2=2π×10−6 TB_2 = 2\pi \times 10^{-6} \, \text{T}
Step 3: Net Magnetic Field
The two coils have currents in opposite directions, so their magnetic fields subtract:
Bnet=B2−B1B_{\text{net}} = B_2 - B_1
Substitute B2=2π×10−6 TB_2 = 2\pi \times 10^{-6} \, \text{T} and B1=2π×10−6 TB_1 = 2\pi \times 10^{-6} \, \text{T}:
Bnet=2π×10−6−2π×10−6B_{\text{net}} = 2\pi \times 10^{-6} - 2\pi \times 10^{-6} Bnet=0 TB_{\text{net}} = 0 \, \text{T}
The net magnetic induction at the center is:
0 T.0 \, \text{T}.
