Calculate the couple (in μ N m ) acting on a magnet of length 10 cm and pole strength 15 A m , kept in a field of B=2× 10- 5 T at an angle of 30° .

Question

Calculate the couple (in μ N m ) acting on a magnet of length 10 cm and pole strength 15 A m , kept in a field of B=2× 10- 5 T at an angle of 30° . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

C=MBsin θ =(m × 2 l)× 2× (10)- 5sin 30° =15× 10× 10- 2× 2× 10- 5× (1/2) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ijivupc7ebiw.png" /> =15× 10- 6 Nm

Q. Calculate the couple (in $μ N m$ ) acting on a magnet of length $10 c m$ and pole strength $15 A m$ , kept in a field of $B = 2 \times 1 0^{- 5} T$ at an angle of $3 0^{\circ}$ .

$C = MB s in θ$
$= (m \times 2 l) \times 2 \times (10)^{- 5} s in 3 0^{\circ}$
$= 15 \times 10 \times 1 0^{- 2} \times 2 \times 1 0^{- 5} \times \frac{1}{2}$

$= 15 \times 1 0^{- 6} N m$

Q. Calculate the couple (in μNm ) acting on a magnet of length 10cm and pole strength 15Am , kept in a field of B=2×10−5T at an angle of 30∘ .

C=MBsinθ =(m×2l)×2×(10)−5sin30∘ =15×10×10−2×2×10−5×21​ =15×10−6Nm

Q. Calculate the couple (in $μ N m$ ) acting on a magnet of length $10 c m$ and pole strength $15 A m$ , kept in a field of $B = 2 \times 1 0^{- 5} T$ at an angle of $3 0^{\circ}$ .

$C = MB s in θ$
$= (m \times 2 l) \times 2 \times (10)^{- 5} s in 3 0^{\circ}$
$= 15 \times 10 \times 1 0^{- 2} \times 2 \times 1 0^{- 5} \times \frac{1}{2}$

$= 15 \times 1 0^{- 6} N m$