Solveeit Logo
Login

Question

Question: A closely wound solenoid of 2000 turns and area of cross-section \( 1.5 \times {10^{ - 4}}{m^2} \) c...

A closely wound solenoid of 2000 turns and area of cross-section 1.5×10−4m21.5 \times {10^{ - 4}}{m^2} carries a current of 2.0A2.0A . It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5×10−2 tesla5 \times {10^{ - 2}}{\text{ }}tesla making an angle of 30∘{30^\circ } with the axis of the solenoid. The torque on the solenoid will be:
(A) 3×10−3Nm3 \times {10^{ - 3}}Nm
(B) 1.5×10−3Nm1.5 \times {10^{ - 3}}Nm
(C) 1.5×10−2Nm1.5 \times {10^{ - 2}}Nm
(D) 3×10−2Nm3 \times {10^{ - 2}}Nm

Explanation

Solution

Hint : In order to answer this question, first we will rewrite the given facts as per the question, and then we will need to find the magnetic moment by applying its formula to find the torque applied on the solenoid. Then we will apply the formula of torque in the terms of magnetic moment and magnetic field, i.e. .

Complete Step By Step Answer:
As per the question-
The number of turns of solenoid is N=2000N = 2000 .
Area of cross-section, A=1.5×10−4m2A = 1.5 \times {10^{ - 4}}{m^2}
Current carrying in a loop of coil, A=2.0AA = 2.0A .
Magnetic field, B=5×10−2teslaB = 5 \times {10^{ - 2}}tesla
And the uniform magnetic field making an angle of, θ=30∘\theta = {30^\circ } .
Now, as we know that-
On current-carrying loop, the torque applied is:
T=m⃗×B⃗T = \vec{m} \times \vec{B}
where, m⃗\vec{m} is the magnetic moment and,
B⃗\vec{B} is the magnetic field of the coil.
⇒T=mBsin⁡θ\Rightarrow T = mB\sin \theta
Now, to find the torque, we need to first find the magnetic moment:
So, we will apply the formula of magnetic moment:
∴m=NIA\therefore m = NIA
where, NN is the number of turns of solenoid,
II is the current carrying by the loop and,
AA is the cross-sectional area of the coil of solenoid.
⇒m=2000×2×1.5×10−4 ⇒m=0.6JT−1\Rightarrow m = 2000 \times 2 \times 1.5 \times {10^{ - 4}} \\\ \Rightarrow m = 0.6J{T^{ - 1}}
Now, we have the value of both magnetic moment and the magnetic field, so we can easily find the value of torque applied:
∴T=mBsin⁡θ         =0.6×5×10−2×sin⁡30∘         =1.5×10−2Nm\therefore T = mB\sin \theta \\\ \,\,\,\,\,\,\,\, = 0.6 \times 5 \times {10^{ - 2}} \times \sin {30^\circ } \\\ \,\,\,\,\,\,\,\, = 1.5 \times {10^{ - 2}}Nm
Therefore, the required torque on the solenoid is 1.5×10−2Nm1.5 \times {10^{ - 2}}Nm .
Hence, the correct option is (C) 1.5×10−2Nm1.5 \times {10^{ - 2}}Nm .

Note :
As if it were a flat single-turn coil, each turn will be torqued about a cross-sectional diameter perpendicular to the B direction. A force also exists in the direction perpendicular to B and the solenoid axis.