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Question: A coil \(10cm\) diameter has \(300turns\). If the coil carries a current of \(10mA\) and lies in a m...

A coil 10cm10cm diameter has 300turns300turns. If the coil carries a current of 10mA10mA and lies in a magnetic field 5×10−2T5 \times {10^{ - 2}}T, the maximum torque experienced by the coil is:
A) 4.7×10−2Nm4.7 \times {10^{ - 2}}Nm
B) 4.7×10−4Nm4.7 \times {10^{ - 4}}Nm
C) 4.7×10−5Nm4.7 \times {10^{ - 5}}Nm
D) 4.7×10−8Nm4.7 \times {10^{ - 8}}Nm

Explanation

Solution

When a current flows inside the coil it generates a magnetic flux around the coal, now as the body is placed between a magnetic field, the magnetic field of the body and magnetic field in which it has been placed opposes each other and the body experiences a torque and tends to rotate.

Complete step by step solution:
The torque experienced by a current-carrying coil depends upon the current flowing through it, the number of turns, the area of the coil, and the magnetic field in which it is placed so,
∴τ=BIAsin⁡θ×n\therefore \tau = BIA\sin \theta \times n
BB is the magnetic field in which the coil is placed
II is the current flowing through the coil
AA is the cross-sectional area of the coil
θ\theta is the angle at which the coil is placed
nn the number of turns in the coil
Here it is asked for the maximum torque and we know the maximum value onsin⁡θ=1\sin \theta = 1at θ=90∘\theta = 90^\circ So the coil should make an angle of90∘90^\circ with magnetic field
Now the various given values are
B=5×10−2TB = 5 \times {10^{ - 2}}T
I=10mAI = 10mA the value of current should be in ampere so we convert the value into amperes
∵1A=1000mA\because 1A = 1000mA
∴10mA=0.01A\therefore 10mA = 0.01A
A=πr2A = \pi {r^2}
Here diameter is given as 2cm2cm so the radius will be 1cm1cmbut we need to convert this into the meter so, r=0.01m r = 0.01m
∴A=π×(0.01)2\therefore A = \pi \times {\left( {0.01} \right)^2}
⇒A=3.14×10−4\Rightarrow A = 3.14 \times {10^{ - 4}}
Substituting all the values in the equation we get

τ=(5×10−2)×0.01×(3.14×10−4)sin⁡90∘×300 ⇒τ=4.71×10−5Nm  \tau = \left( {5 \times {{10}^{ - 2}}} \right) \times 0.01 \times \left( {3.14 \times {{10}^{ - 4}}} \right)\sin 90^\circ \times 300 \\\ \Rightarrow \tau = 4.71 \times {10^{ - 5}}Nm \\\

Final answer is (C), the total torque experienced by the coil will be 4.71×101−5Nm.4.71 \times {101^{ - 5}}Nm.

Note: When the current flows through any conductor it generates a magnetic field around it. The magnetic field of the conductor when it comes in between other magnetic fields it experiences a torque which tends to rotate it, this is called the magnetic field of electric current. This is the explanation of what happens inside an electric motor.
The torque will increase by increasing the no of turns or the current passing through the coil.