NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII04:MOVING CHARGES AND MAGNETISM
362669
A vertical straight conductor carries a current upward. A point \(P\) lies to the East of it at a small distance and another point \(Q\) lies to the West at the same distance. The magnetic field at \(P\) is
1 greater than at \(Q\)
2 same as at \(Q\)
3 less than at \(Q\)
4 greater or less than at \(Q\) depending upon the strength of current
Explanation:
Magnitude of magnetic field at any point due to a straight current carrying conductor depends only on the distance of the point with respect to current carrying conductor but not on the direction. Hence, \(B_{P}=B_{Q}\) (at same distance).
PHXII04:MOVING CHARGES AND MAGNETISM
362670
Assertion : When current is represented by a straight line, the magnetic field will be circular. Reason : According to Fleming's left hand rule, direction of force is parallel to the magnetic field.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
So correct option is (3).
PHXII04:MOVING CHARGES AND MAGNETISM
362671
The magnetic field at the centre of a circular coil of radius \({R}\) carrying current \({I}\) is 64 times the magnetic field at a distance \({x}\) on its axis from the centre of the coil. Then the value of \({x}\) is
362672
A circular coil carrying current \('I'\) has radius ' \(R\) ' and magnetic field at the centre is ' \(B\) '. At what distance from the centre along the axis of the same coil, the magnetic field will be \(\dfrac{B}{8}\) ?
1 \(R \sqrt{2}\)
2 \(R \sqrt{3}\)
3 2\(R\)
4 3\(R\)
Explanation:
According to the question, Magnetic field at the centre of circular coil is given by \(B_{\text {centre }}=\dfrac{\mu_{0} N i}{2 R}\) \(B_{a x i s}=\dfrac{\mu_{0}}{4 \pi} \dfrac{2 \pi N i R^{2}}{\left(x^{2}+R^{2}\right)^{3 / 2}}\) \(\dfrac{B}{8}=\dfrac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}}\) From eq.(1), we get \(\begin{aligned}& \dfrac{\mu_{0} N i}{8 \times 2 R}=\dfrac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} \\& \dfrac{1}{8 R^{3}}=\dfrac{1}{\left(x^{2}+R^{2}\right)^{3 / 2}} \\& 8 R^{3}=\left(x^{2}+R^{2}\right)^{3 / 2} \\& 2 R=\left(x^{2}+R^{2}\right)^{1 / 2}(\text { on squaring) } \\& 4 R^{2}=\left(x^{2}+R^{2}\right) \\& 3 R^{2}=x^{2} \Rightarrow x=R \sqrt{3}\end{aligned}\)
362669
A vertical straight conductor carries a current upward. A point \(P\) lies to the East of it at a small distance and another point \(Q\) lies to the West at the same distance. The magnetic field at \(P\) is
1 greater than at \(Q\)
2 same as at \(Q\)
3 less than at \(Q\)
4 greater or less than at \(Q\) depending upon the strength of current
Explanation:
Magnitude of magnetic field at any point due to a straight current carrying conductor depends only on the distance of the point with respect to current carrying conductor but not on the direction. Hence, \(B_{P}=B_{Q}\) (at same distance).
PHXII04:MOVING CHARGES AND MAGNETISM
362670
Assertion : When current is represented by a straight line, the magnetic field will be circular. Reason : According to Fleming's left hand rule, direction of force is parallel to the magnetic field.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
So correct option is (3).
PHXII04:MOVING CHARGES AND MAGNETISM
362671
The magnetic field at the centre of a circular coil of radius \({R}\) carrying current \({I}\) is 64 times the magnetic field at a distance \({x}\) on its axis from the centre of the coil. Then the value of \({x}\) is
362672
A circular coil carrying current \('I'\) has radius ' \(R\) ' and magnetic field at the centre is ' \(B\) '. At what distance from the centre along the axis of the same coil, the magnetic field will be \(\dfrac{B}{8}\) ?
1 \(R \sqrt{2}\)
2 \(R \sqrt{3}\)
3 2\(R\)
4 3\(R\)
Explanation:
According to the question, Magnetic field at the centre of circular coil is given by \(B_{\text {centre }}=\dfrac{\mu_{0} N i}{2 R}\) \(B_{a x i s}=\dfrac{\mu_{0}}{4 \pi} \dfrac{2 \pi N i R^{2}}{\left(x^{2}+R^{2}\right)^{3 / 2}}\) \(\dfrac{B}{8}=\dfrac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}}\) From eq.(1), we get \(\begin{aligned}& \dfrac{\mu_{0} N i}{8 \times 2 R}=\dfrac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} \\& \dfrac{1}{8 R^{3}}=\dfrac{1}{\left(x^{2}+R^{2}\right)^{3 / 2}} \\& 8 R^{3}=\left(x^{2}+R^{2}\right)^{3 / 2} \\& 2 R=\left(x^{2}+R^{2}\right)^{1 / 2}(\text { on squaring) } \\& 4 R^{2}=\left(x^{2}+R^{2}\right) \\& 3 R^{2}=x^{2} \Rightarrow x=R \sqrt{3}\end{aligned}\)
362669
A vertical straight conductor carries a current upward. A point \(P\) lies to the East of it at a small distance and another point \(Q\) lies to the West at the same distance. The magnetic field at \(P\) is
1 greater than at \(Q\)
2 same as at \(Q\)
3 less than at \(Q\)
4 greater or less than at \(Q\) depending upon the strength of current
Explanation:
Magnitude of magnetic field at any point due to a straight current carrying conductor depends only on the distance of the point with respect to current carrying conductor but not on the direction. Hence, \(B_{P}=B_{Q}\) (at same distance).
PHXII04:MOVING CHARGES AND MAGNETISM
362670
Assertion : When current is represented by a straight line, the magnetic field will be circular. Reason : According to Fleming's left hand rule, direction of force is parallel to the magnetic field.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
So correct option is (3).
PHXII04:MOVING CHARGES AND MAGNETISM
362671
The magnetic field at the centre of a circular coil of radius \({R}\) carrying current \({I}\) is 64 times the magnetic field at a distance \({x}\) on its axis from the centre of the coil. Then the value of \({x}\) is
362672
A circular coil carrying current \('I'\) has radius ' \(R\) ' and magnetic field at the centre is ' \(B\) '. At what distance from the centre along the axis of the same coil, the magnetic field will be \(\dfrac{B}{8}\) ?
1 \(R \sqrt{2}\)
2 \(R \sqrt{3}\)
3 2\(R\)
4 3\(R\)
Explanation:
According to the question, Magnetic field at the centre of circular coil is given by \(B_{\text {centre }}=\dfrac{\mu_{0} N i}{2 R}\) \(B_{a x i s}=\dfrac{\mu_{0}}{4 \pi} \dfrac{2 \pi N i R^{2}}{\left(x^{2}+R^{2}\right)^{3 / 2}}\) \(\dfrac{B}{8}=\dfrac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}}\) From eq.(1), we get \(\begin{aligned}& \dfrac{\mu_{0} N i}{8 \times 2 R}=\dfrac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} \\& \dfrac{1}{8 R^{3}}=\dfrac{1}{\left(x^{2}+R^{2}\right)^{3 / 2}} \\& 8 R^{3}=\left(x^{2}+R^{2}\right)^{3 / 2} \\& 2 R=\left(x^{2}+R^{2}\right)^{1 / 2}(\text { on squaring) } \\& 4 R^{2}=\left(x^{2}+R^{2}\right) \\& 3 R^{2}=x^{2} \Rightarrow x=R \sqrt{3}\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII04:MOVING CHARGES AND MAGNETISM
362669
A vertical straight conductor carries a current upward. A point \(P\) lies to the East of it at a small distance and another point \(Q\) lies to the West at the same distance. The magnetic field at \(P\) is
1 greater than at \(Q\)
2 same as at \(Q\)
3 less than at \(Q\)
4 greater or less than at \(Q\) depending upon the strength of current
Explanation:
Magnitude of magnetic field at any point due to a straight current carrying conductor depends only on the distance of the point with respect to current carrying conductor but not on the direction. Hence, \(B_{P}=B_{Q}\) (at same distance).
PHXII04:MOVING CHARGES AND MAGNETISM
362670
Assertion : When current is represented by a straight line, the magnetic field will be circular. Reason : According to Fleming's left hand rule, direction of force is parallel to the magnetic field.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
So correct option is (3).
PHXII04:MOVING CHARGES AND MAGNETISM
362671
The magnetic field at the centre of a circular coil of radius \({R}\) carrying current \({I}\) is 64 times the magnetic field at a distance \({x}\) on its axis from the centre of the coil. Then the value of \({x}\) is
362672
A circular coil carrying current \('I'\) has radius ' \(R\) ' and magnetic field at the centre is ' \(B\) '. At what distance from the centre along the axis of the same coil, the magnetic field will be \(\dfrac{B}{8}\) ?
1 \(R \sqrt{2}\)
2 \(R \sqrt{3}\)
3 2\(R\)
4 3\(R\)
Explanation:
According to the question, Magnetic field at the centre of circular coil is given by \(B_{\text {centre }}=\dfrac{\mu_{0} N i}{2 R}\) \(B_{a x i s}=\dfrac{\mu_{0}}{4 \pi} \dfrac{2 \pi N i R^{2}}{\left(x^{2}+R^{2}\right)^{3 / 2}}\) \(\dfrac{B}{8}=\dfrac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}}\) From eq.(1), we get \(\begin{aligned}& \dfrac{\mu_{0} N i}{8 \times 2 R}=\dfrac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} \\& \dfrac{1}{8 R^{3}}=\dfrac{1}{\left(x^{2}+R^{2}\right)^{3 / 2}} \\& 8 R^{3}=\left(x^{2}+R^{2}\right)^{3 / 2} \\& 2 R=\left(x^{2}+R^{2}\right)^{1 / 2}(\text { on squaring) } \\& 4 R^{2}=\left(x^{2}+R^{2}\right) \\& 3 R^{2}=x^{2} \Rightarrow x=R \sqrt{3}\end{aligned}\)
