A toroid has 500 turns per metre length. If it carries a current of 2A, the magnetic energy density inside the toroid is

Question

A toroid has 500 turns per metre length. If it carries a current of 2A, the magnetic energy density inside the toroid is (A) 6.28 J/m3 (B) 0.628 J/m3 (C) 3.14 J/m3 (D) 0.314 J/m3

Tardigrade Admin · Accepted Answer

Magnetic field inside a toroid B=μ0 nl Magnetic energy density inside the toroid is u0=(B2/2 μ0)=(μ 0 2n2 I2/2 μ0)=(μ0n2I2/2) =(4π×10-7×(500)2×22/2)=0.628 J m-3

Q. A toroid has 500 turns per metre length. If it carries a current of 2A, the magnetic energy density inside the toroid is

$6.28 J / m^{3}$

$0.628 J / m^{3}$

$3.14 J / m^{3}$

$0.314 J / m^{3}$

Magnetic field inside a toroid
$B = μ_{0} n l$
Magnetic energy density inside the toroid is
$u_{0} = \frac{B ^{2}}{2 μ _{0}} = \frac{μ _{0}^{2} n ^{2} I ^{2}}{2 μ _{0}} = \frac{μ _{0} n ^{2} I ^{2}}{2}$
$= \frac{4 π \times 1 0 ^{- 7} \times ( 500 ) ^{2} \times 2 ^{2}}{2} = 0.628 J m^{- 3}$

Q. A toroid has 500 turns per metre length. If it carries a current of 2A, the magnetic energy density inside the toroid is

6.28J/m3

0.628J/m3

3.14J/m3

0.314J/m3

Magnetic field inside a toroid B=μ0​nl Magnetic energy density inside the toroid is u0​=2μ0​B2​=2μ0​μ02​n2I2​=2μ0​n2I2​ =24π×10−7×(500)2×22​=0.628Jm−3

$6.28 J / m^{3}$

$0.628 J / m^{3}$

$3.14 J / m^{3}$

$0.314 J / m^{3}$

Magnetic field inside a toroid
$B = μ_{0} n l$
Magnetic energy density inside the toroid is
$u_{0} = \frac{B ^{2}}{2 μ _{0}} = \frac{μ _{0}^{2} n ^{2} I ^{2}}{2 μ _{0}} = \frac{μ _{0} n ^{2} I ^{2}}{2}$
$= \frac{4 π \times 1 0 ^{- 7} \times ( 500 ) ^{2} \times 2 ^{2}}{2} = 0.628 J m^{- 3}$