If a magnet of length $10{\text{ }}cm$ and pole strength $40A - m$ is placed at an angle of ${45^ \circ }$ in a uniform induction field of intensity $2 \times {10^{ - 4}}T$ . The couple acting on it is
A. $0.5656 \times {10^{ - 4}}N - m$
B. $0.565 \times {10^{ - 3}}N - m$
C. $0.756 \times {10^{ - 4}}N - m$
D. None of the above
Answer
251.7k+ views
Hint:
In the case of a problem based on a magnetic field, we know that the couple acting on a magnet directly varies with the magnetic field. Also, we know that there is a relation between the magnetic field and torque generated in bar magnet i.e., $\tau = \overrightarrow M \times \overrightarrow B $ which we will use to get the solution to the given problem.
Complete step by step solution:
A magnet of length $l = 10cm = 0.1m$ and pole strength $m = 40A - m$ is placed in a uniform induction field of intensity, $B = 2 \times {10^{ - 4}}T$ (given)
A couple of moment i.e., torque $\tau $is experienced by a magnet, when it is placed at an angle of $\theta = {45^ \circ }$
We know that the formula for the torque acting on a magnet in a uniform magnetic field is given as:
$\tau = \overrightarrow M \times \overrightarrow B $
or, $\tau = MB\sin \theta $ … (1)
Here $M = ml$ is magnetic dipole moment
From eq. (1)
$ \Rightarrow \tau = MB\sin \theta = mlB\sin \theta $
Substituting all the required values from the question in the above expression, we get
$ \Rightarrow \tau = 40 \times 0.1 \times \left( {2 \times {{10}^{ - 4}}} \right) \times \sin ({45^ \circ })$
$ \Rightarrow \tau = 8 \times {10^{ - 4}} \times 0.7071 = 5.656 \times {10^{ - 4}}$ $\left( {\therefore \sin {{45}^ \circ } = \dfrac{1}{{\sqrt 2 }} = 0.7071} \right)$
Approximating this value, we get
$ \Rightarrow \tau = 0.565 \times {10^{ - 3}}N - m$
Thus, the couple acting on a magnet is $0.565 \times {10^{ - 3}}N - m$.
Hence, the correct option is (B) $0.565 \times {10^{ - 3}}N - m$ .
Therefore, the correct option is B.
Note:
Since this is a problem related to uniform magnetic field and torque, quantities required to calculate the couple acting on a magnet must be identified on a prior basis as it gives a better understanding of the problem and helps to solve the question further. Units must be there after each physical quantity.
In the case of a problem based on a magnetic field, we know that the couple acting on a magnet directly varies with the magnetic field. Also, we know that there is a relation between the magnetic field and torque generated in bar magnet i.e., $\tau = \overrightarrow M \times \overrightarrow B $ which we will use to get the solution to the given problem.
Complete step by step solution:
A magnet of length $l = 10cm = 0.1m$ and pole strength $m = 40A - m$ is placed in a uniform induction field of intensity, $B = 2 \times {10^{ - 4}}T$ (given)
A couple of moment i.e., torque $\tau $is experienced by a magnet, when it is placed at an angle of $\theta = {45^ \circ }$
We know that the formula for the torque acting on a magnet in a uniform magnetic field is given as:
$\tau = \overrightarrow M \times \overrightarrow B $
or, $\tau = MB\sin \theta $ … (1)
Here $M = ml$ is magnetic dipole moment
From eq. (1)
$ \Rightarrow \tau = MB\sin \theta = mlB\sin \theta $
Substituting all the required values from the question in the above expression, we get
$ \Rightarrow \tau = 40 \times 0.1 \times \left( {2 \times {{10}^{ - 4}}} \right) \times \sin ({45^ \circ })$
$ \Rightarrow \tau = 8 \times {10^{ - 4}} \times 0.7071 = 5.656 \times {10^{ - 4}}$ $\left( {\therefore \sin {{45}^ \circ } = \dfrac{1}{{\sqrt 2 }} = 0.7071} \right)$
Approximating this value, we get
$ \Rightarrow \tau = 0.565 \times {10^{ - 3}}N - m$
Thus, the couple acting on a magnet is $0.565 \times {10^{ - 3}}N - m$.
Hence, the correct option is (B) $0.565 \times {10^{ - 3}}N - m$ .
Therefore, the correct option is B.
Note:
Since this is a problem related to uniform magnetic field and torque, quantities required to calculate the couple acting on a magnet must be identified on a prior basis as it gives a better understanding of the problem and helps to solve the question further. Units must be there after each physical quantity.
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