Chapter 30: Q66P (page 900)

A circular loop of wire 50 mmin radius carries a current of 100 A. (a) Find the magnetic field strength. (b) Find the energy density at the center of the loop

Short Answer

Expert verified

$a) B = 1.3 \times 10^{- 3} T b) u_{B} = 0.63 \frac{J}{m^{3}}$

Step by step solution

Given

i) Radius of the circular loop $R = 50 mm = 50 \times 10^{- 3} m$

ii) Current in the circular loop $i = 100 A$

Understanding the concept

We can use the formula of magnetic field at the center of the current loop to find it. Using this magnetic field into the formula of the energy density at that point, we can find the energy density at the center of the loop.

Formula:

$B = μ_{0} i / 2 R u_{B} = B^{2} / 2 μ_{0}$

(a) Calculate the magnetic field strength

The magnetic field at the center of loop is given by

$B = \frac{μ_{0} i}{2 R} B = \frac{(4 \times 10^{- 7} \frac{H}{m}) (100 A)}{2 (50 \times 10^{- 3} m)} B = 1.3 \times 10^{- 3} T$

Therefore, the magnetic field strength is $1.3 \times 10^{- 3} T$ .

(b) Calculate the energy density at the center of the loop

The energy density at the centre of the loop is given by

$u_{B} = \frac{B^{2}}{2 μ_{0}} u_{B} = \frac{{(1.3 \times 10^{- 3} T)}^{2}}{2 (4 π \times 10^{- 7} \frac{H}{m})} u_{B} = 0.63 \frac{J}{m^{3}}$

Therefore, energy density at the center of the loop is role="math" localid="1661424939827" $0.63 J / m^{3}$ .

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A circular loop of wire 50 mmin radius carries a current of 100 A. (a) Find the magnetic field strength. (b) Find the energy density at the center of the loop

Short Answer

Step by step solution

Given

Understanding the concept

(a) Calculate the magnetic field strength

(b) Calculate the energy density at the center of the loop

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