A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T

Question

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of $30 °$ with the direction of the magnetic field. The torque acting on the coil is $α \times 10^{- 4} Nm$ . The value of $α$ is ________. $(π = 3.14)$ [2026]

Smart Achievers · Accepted Answer

(314)

$τ =$ $| \vec{M} \times \vec{B} |$ $=$ $| N I \vec{A} \times \vec{B} |$

$τ = 125 \times 1 \times π {(\frac{2}{100})}^{2} \times 0.4 \times \sin 30 °$

$= 100 π \times 10^{- 4} N m$

$So α = 100 π = 314$

Solution

Ans.
(314)

$τ =$ $| \vec{M} \times \vec{B} |$ $=$ $| N I \vec{A} \times \vec{B} |$

$τ = 125 \times 1 \times π {(\frac{2}{100})}^{2} \times 0.4 \times \sin 30 °$

$= 100 π \times 10^{- 4} N m$

$So α = 100 π = 314$

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Solution

Ans. (314) τ=|M→×B→|=|NIA→×B→| τ=125×1×π(2100)2×0.4×sin30° =100π×10-4 N m So α=100π=314

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Ans.
(314)

$τ =$ $| \vec{M} \times \vec{B} |$ $=$ $| N I \vec{A} \times \vec{B} |$

$τ = 125 \times 1 \times π {(\frac{2}{100})}^{2} \times 0.4 \times \sin 30 °$

$= 100 π \times 10^{- 4} N m$

$So α = 100 π = 314$