
A long straight wire in the horizontal plane carries a current of 50A in north to south direction. Find the magnitude and direction of B at a point 2.5m east of the wire.
Answer
600.3k+ views
Hint: We will first apply the formula for magnetic field for an infinite conductor. We will substitute the value of ${\mu _0}$ in the formula including the other values of current and radius as per given in the question. Then we will apply the right-hand rule to know the direction. Refer to the solution below.
Complete step-by-step solution -
Formula used: $B = \dfrac{{{\mu _0}I}}{{2\pi r}}$.
As we already know that for an infinite conductor, the magnetic field $B$ at a perpendicular distance r is equal to-
$ \Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi r}}$
As we know that ${\mu _0}$ is equal to $4\pi \times {10^{ - 7}}$. We will substitute this value into the equation above as well as the values of current $I$, and the value of radius-
Current $I$ as per given in the question = 50A
Radius r as per given in the question = 2.5m
Thus, we will have-
$ \Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi r}} \\
\Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 50}}{{2\pi \times 2.5}} \\
\Rightarrow B = \dfrac{{2 \times {{10}^{ - 7}} \times 500}}{{25}} \\
\Rightarrow B = 4 \times {10^{ - 6}}T \\ $
Now, we know that the magnetic field has a value of $4 \times {10^{ - 6}}T$.
We will use the right-hand rule, pointing the thumb in the direction of the current and our fingers will curl in the direction of the magnetic field. We will notice that at point P, the magnetic field will be out of the plane of the paper or will be a dot field. Hence, it acts upwards which is perpendicular to the plane of the wire.
Note: The right-hand rule in mathematics and physics is a popular mnemonic for recognizing three-dimensional axis orientation. You can see by holding your hands out and palms up, curling your fingers, and the thumb extending. The third or z-axis will point to each thumb if the curl of the fingers indicates a shift from the main or x-axis to the second or y-axis.
Complete step-by-step solution -
Formula used: $B = \dfrac{{{\mu _0}I}}{{2\pi r}}$.
As we already know that for an infinite conductor, the magnetic field $B$ at a perpendicular distance r is equal to-
$ \Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi r}}$
As we know that ${\mu _0}$ is equal to $4\pi \times {10^{ - 7}}$. We will substitute this value into the equation above as well as the values of current $I$, and the value of radius-
Current $I$ as per given in the question = 50A
Radius r as per given in the question = 2.5m
Thus, we will have-
$ \Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi r}} \\
\Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 50}}{{2\pi \times 2.5}} \\
\Rightarrow B = \dfrac{{2 \times {{10}^{ - 7}} \times 500}}{{25}} \\
\Rightarrow B = 4 \times {10^{ - 6}}T \\ $
Now, we know that the magnetic field has a value of $4 \times {10^{ - 6}}T$.
We will use the right-hand rule, pointing the thumb in the direction of the current and our fingers will curl in the direction of the magnetic field. We will notice that at point P, the magnetic field will be out of the plane of the paper or will be a dot field. Hence, it acts upwards which is perpendicular to the plane of the wire.
Note: The right-hand rule in mathematics and physics is a popular mnemonic for recognizing three-dimensional axis orientation. You can see by holding your hands out and palms up, curling your fingers, and the thumb extending. The third or z-axis will point to each thumb if the curl of the fingers indicates a shift from the main or x-axis to the second or y-axis.
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