First slide

Magnetic field due to current element

Question

A solenoid of 0.4m length with 500 turns carries a current of 3A. A coil of 10 turns and of radius 0.01m carries a current of 0.4A. The torque required to hold the coil with its axis at right angles to that of solenoid in the middle point of it is,

Moderate

A

$6 π^{2} \times 10^{- 7} Nm$
B

$3 π^{2} \times 10^{- 7} Nm$
C

$9 π^{2} \times 10^{- 7} Nm$
D

$12 π^{2} \times 10^{- 7} Nm$

Solution

Field due to solenoid, $B = μ_{o} n I = 4 π \times 10^{- 7} \times \frac{500}{0.4} \times 3 = 15 π \times 10^{- 4} T$
Now, torque experienced by coil in presence of above field $Torque = τ = B N i A = B N i (π r^{2}) = 15 π \times 10^{- 4} \times 10 \times 0.4 \times π \times {(0.01)}^{2} = 6 π^{2} \times 10^{- 7} Nm$

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