154032
A straight wire carrying current $I$ is made into a circular loop. If $M$ is the magnetic moment associated with the loop, then the length of the wire is
1 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
2 $\sqrt{\frac{2 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
4 $\sqrt{\frac{\pi \mathrm{M}}{2 \mathrm{I}}}$
5 $\sqrt{\frac{\pi \mathrm{M}}{4 \mathrm{I}}}$
Explanation:
A Let, the length of wire $=\mathrm{L}$ When, wire is bent in the form of circle. Then, circumference of circle $=$ length of wire $2 \pi \mathrm{r}=\mathrm{L}$ Where, $r=$ Radius of circle Then, $r=\frac{L}{2 \pi}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=\pi \times\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}=\frac{\pi \mathrm{L}^{2}}{4 \pi^{2}}=\frac{\mathrm{L}^{2}}{4 \pi}$ We know that, Magnetic moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M} =\mathrm{I} \times \frac{\mathrm{L}^{2}}{4 \pi}$ $\mathrm{L}^{2} =\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\text { So, } \quad \mathrm{L} =\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
Kerala CEE 2012
Magnetism and Matter
154039
The effective length of a magnet is $31.4 \mathrm{~cm}$ and its pole strength is $0.8 \mathrm{Am}$. The magnetic moment, if it is bent in the form of a semicircle is $\mathbf{A m}^{2}$
1 0.12
2 1.2
3 0.16
4 1.6
Explanation:
C Given, Effective length wire $(l)=31.4 \mathrm{~cm}=31.4 \times 10^{-2} \mathrm{~m}$ Pole strength $(\mathrm{m})=0.8 \mathrm{Am}$ When magnet is bent inform of semicircle - Then, circumference of semicircle $=$ length of magnet $\pi \mathrm{r}=31.4 \times 10^{-2}$ $\operatorname{Radius}(\mathrm{r})=\frac{31.4 \times 10^{-2}}{\pi}$ Diameter $=2 \mathrm{r}$ Initially magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{M}=0.8 \times 31.4 \times 10^{-2}$ $\mathrm{M}=25.12 \times 10^{-2}$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime}=0.8 \times 2 \times \frac{31.4 \times 10^{-2}}{3.14}$ $\mathrm{M}^{\prime}=0.16 \mathrm{Am}^{2}$
GUJCET 2016
Magnetism and Matter
154040
A bar magnet of magnetic moment $M$ is placed at right angles to a magnetic induction $B$. If a force $F$ is experienced by each pole of the magnet, the length of the magnet will be
1 $\mathrm{MB} / \mathrm{F}$
2 $\mathrm{BF} / \mathrm{M}$
3 $\mathrm{MF} / \mathrm{B}$
4 $\mathrm{F} / \mathrm{MB}$
Explanation:
A Given, Magnetic moment $=\mathrm{M}$ Magnetic induction $=\mathrm{B}, \theta=90^{\circ}$ We know that, $\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$ $\tau=\mathrm{MB} \sin \theta$ $\tau=\mathrm{MB} \sin 90^{\circ}$ $\tau=\mathrm{MB}$ The torque experienced by magnet- $\tau=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$ $\tau=\mathrm{rF} \sin 90^{\circ}$ $\tau=\mathrm{rF}$ Here, $\mathrm{r}=l$ ( $l=$ length of magnet) $\therefore \quad \tau=l \mathrm{~F}$ From equation (i) and (ii), we get- $l \mathrm{~F}=\mathrm{MB}$ $l=\frac{\mathrm{MB}}{\mathrm{F}}$
COMEDK 2013
Magnetism and Matter
154042
Assertion: A disc-shaped magnet is deviated above a superconducting material that has been cooled by liquid nitrogen. Reason: Super conductors repel a magnet.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A A superconductor are diamagnetic in nature. So, when it is placed above a magnet it will repelled by magnet or it will be move to from higher field to lower field. This is the principle behind deviation of superconductor above magnet. Superconductor exist only a below a certain temperature.
AIIMS-2005
Magnetism and Matter
154041
Assertion: Electromagnetic are made of soft iron. Reason: Coercivity of soft iron is small.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B As the material of electromagnet is subjected to magnetize and demagnetize, the hysteresis loss should be small. Material should gain high value I and $B$ with low value of magnetic field intensity. So the soft iron is preferred. Whose coercively is small.
154032
A straight wire carrying current $I$ is made into a circular loop. If $M$ is the magnetic moment associated with the loop, then the length of the wire is
1 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
2 $\sqrt{\frac{2 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
4 $\sqrt{\frac{\pi \mathrm{M}}{2 \mathrm{I}}}$
5 $\sqrt{\frac{\pi \mathrm{M}}{4 \mathrm{I}}}$
Explanation:
A Let, the length of wire $=\mathrm{L}$ When, wire is bent in the form of circle. Then, circumference of circle $=$ length of wire $2 \pi \mathrm{r}=\mathrm{L}$ Where, $r=$ Radius of circle Then, $r=\frac{L}{2 \pi}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=\pi \times\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}=\frac{\pi \mathrm{L}^{2}}{4 \pi^{2}}=\frac{\mathrm{L}^{2}}{4 \pi}$ We know that, Magnetic moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M} =\mathrm{I} \times \frac{\mathrm{L}^{2}}{4 \pi}$ $\mathrm{L}^{2} =\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\text { So, } \quad \mathrm{L} =\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
Kerala CEE 2012
Magnetism and Matter
154039
The effective length of a magnet is $31.4 \mathrm{~cm}$ and its pole strength is $0.8 \mathrm{Am}$. The magnetic moment, if it is bent in the form of a semicircle is $\mathbf{A m}^{2}$
1 0.12
2 1.2
3 0.16
4 1.6
Explanation:
C Given, Effective length wire $(l)=31.4 \mathrm{~cm}=31.4 \times 10^{-2} \mathrm{~m}$ Pole strength $(\mathrm{m})=0.8 \mathrm{Am}$ When magnet is bent inform of semicircle - Then, circumference of semicircle $=$ length of magnet $\pi \mathrm{r}=31.4 \times 10^{-2}$ $\operatorname{Radius}(\mathrm{r})=\frac{31.4 \times 10^{-2}}{\pi}$ Diameter $=2 \mathrm{r}$ Initially magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{M}=0.8 \times 31.4 \times 10^{-2}$ $\mathrm{M}=25.12 \times 10^{-2}$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime}=0.8 \times 2 \times \frac{31.4 \times 10^{-2}}{3.14}$ $\mathrm{M}^{\prime}=0.16 \mathrm{Am}^{2}$
GUJCET 2016
Magnetism and Matter
154040
A bar magnet of magnetic moment $M$ is placed at right angles to a magnetic induction $B$. If a force $F$ is experienced by each pole of the magnet, the length of the magnet will be
1 $\mathrm{MB} / \mathrm{F}$
2 $\mathrm{BF} / \mathrm{M}$
3 $\mathrm{MF} / \mathrm{B}$
4 $\mathrm{F} / \mathrm{MB}$
Explanation:
A Given, Magnetic moment $=\mathrm{M}$ Magnetic induction $=\mathrm{B}, \theta=90^{\circ}$ We know that, $\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$ $\tau=\mathrm{MB} \sin \theta$ $\tau=\mathrm{MB} \sin 90^{\circ}$ $\tau=\mathrm{MB}$ The torque experienced by magnet- $\tau=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$ $\tau=\mathrm{rF} \sin 90^{\circ}$ $\tau=\mathrm{rF}$ Here, $\mathrm{r}=l$ ( $l=$ length of magnet) $\therefore \quad \tau=l \mathrm{~F}$ From equation (i) and (ii), we get- $l \mathrm{~F}=\mathrm{MB}$ $l=\frac{\mathrm{MB}}{\mathrm{F}}$
COMEDK 2013
Magnetism and Matter
154042
Assertion: A disc-shaped magnet is deviated above a superconducting material that has been cooled by liquid nitrogen. Reason: Super conductors repel a magnet.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A A superconductor are diamagnetic in nature. So, when it is placed above a magnet it will repelled by magnet or it will be move to from higher field to lower field. This is the principle behind deviation of superconductor above magnet. Superconductor exist only a below a certain temperature.
AIIMS-2005
Magnetism and Matter
154041
Assertion: Electromagnetic are made of soft iron. Reason: Coercivity of soft iron is small.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B As the material of electromagnet is subjected to magnetize and demagnetize, the hysteresis loss should be small. Material should gain high value I and $B$ with low value of magnetic field intensity. So the soft iron is preferred. Whose coercively is small.
154032
A straight wire carrying current $I$ is made into a circular loop. If $M$ is the magnetic moment associated with the loop, then the length of the wire is
1 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
2 $\sqrt{\frac{2 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
4 $\sqrt{\frac{\pi \mathrm{M}}{2 \mathrm{I}}}$
5 $\sqrt{\frac{\pi \mathrm{M}}{4 \mathrm{I}}}$
Explanation:
A Let, the length of wire $=\mathrm{L}$ When, wire is bent in the form of circle. Then, circumference of circle $=$ length of wire $2 \pi \mathrm{r}=\mathrm{L}$ Where, $r=$ Radius of circle Then, $r=\frac{L}{2 \pi}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=\pi \times\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}=\frac{\pi \mathrm{L}^{2}}{4 \pi^{2}}=\frac{\mathrm{L}^{2}}{4 \pi}$ We know that, Magnetic moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M} =\mathrm{I} \times \frac{\mathrm{L}^{2}}{4 \pi}$ $\mathrm{L}^{2} =\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\text { So, } \quad \mathrm{L} =\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
Kerala CEE 2012
Magnetism and Matter
154039
The effective length of a magnet is $31.4 \mathrm{~cm}$ and its pole strength is $0.8 \mathrm{Am}$. The magnetic moment, if it is bent in the form of a semicircle is $\mathbf{A m}^{2}$
1 0.12
2 1.2
3 0.16
4 1.6
Explanation:
C Given, Effective length wire $(l)=31.4 \mathrm{~cm}=31.4 \times 10^{-2} \mathrm{~m}$ Pole strength $(\mathrm{m})=0.8 \mathrm{Am}$ When magnet is bent inform of semicircle - Then, circumference of semicircle $=$ length of magnet $\pi \mathrm{r}=31.4 \times 10^{-2}$ $\operatorname{Radius}(\mathrm{r})=\frac{31.4 \times 10^{-2}}{\pi}$ Diameter $=2 \mathrm{r}$ Initially magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{M}=0.8 \times 31.4 \times 10^{-2}$ $\mathrm{M}=25.12 \times 10^{-2}$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime}=0.8 \times 2 \times \frac{31.4 \times 10^{-2}}{3.14}$ $\mathrm{M}^{\prime}=0.16 \mathrm{Am}^{2}$
GUJCET 2016
Magnetism and Matter
154040
A bar magnet of magnetic moment $M$ is placed at right angles to a magnetic induction $B$. If a force $F$ is experienced by each pole of the magnet, the length of the magnet will be
1 $\mathrm{MB} / \mathrm{F}$
2 $\mathrm{BF} / \mathrm{M}$
3 $\mathrm{MF} / \mathrm{B}$
4 $\mathrm{F} / \mathrm{MB}$
Explanation:
A Given, Magnetic moment $=\mathrm{M}$ Magnetic induction $=\mathrm{B}, \theta=90^{\circ}$ We know that, $\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$ $\tau=\mathrm{MB} \sin \theta$ $\tau=\mathrm{MB} \sin 90^{\circ}$ $\tau=\mathrm{MB}$ The torque experienced by magnet- $\tau=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$ $\tau=\mathrm{rF} \sin 90^{\circ}$ $\tau=\mathrm{rF}$ Here, $\mathrm{r}=l$ ( $l=$ length of magnet) $\therefore \quad \tau=l \mathrm{~F}$ From equation (i) and (ii), we get- $l \mathrm{~F}=\mathrm{MB}$ $l=\frac{\mathrm{MB}}{\mathrm{F}}$
COMEDK 2013
Magnetism and Matter
154042
Assertion: A disc-shaped magnet is deviated above a superconducting material that has been cooled by liquid nitrogen. Reason: Super conductors repel a magnet.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A A superconductor are diamagnetic in nature. So, when it is placed above a magnet it will repelled by magnet or it will be move to from higher field to lower field. This is the principle behind deviation of superconductor above magnet. Superconductor exist only a below a certain temperature.
AIIMS-2005
Magnetism and Matter
154041
Assertion: Electromagnetic are made of soft iron. Reason: Coercivity of soft iron is small.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B As the material of electromagnet is subjected to magnetize and demagnetize, the hysteresis loss should be small. Material should gain high value I and $B$ with low value of magnetic field intensity. So the soft iron is preferred. Whose coercively is small.
154032
A straight wire carrying current $I$ is made into a circular loop. If $M$ is the magnetic moment associated with the loop, then the length of the wire is
1 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
2 $\sqrt{\frac{2 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
4 $\sqrt{\frac{\pi \mathrm{M}}{2 \mathrm{I}}}$
5 $\sqrt{\frac{\pi \mathrm{M}}{4 \mathrm{I}}}$
Explanation:
A Let, the length of wire $=\mathrm{L}$ When, wire is bent in the form of circle. Then, circumference of circle $=$ length of wire $2 \pi \mathrm{r}=\mathrm{L}$ Where, $r=$ Radius of circle Then, $r=\frac{L}{2 \pi}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=\pi \times\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}=\frac{\pi \mathrm{L}^{2}}{4 \pi^{2}}=\frac{\mathrm{L}^{2}}{4 \pi}$ We know that, Magnetic moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M} =\mathrm{I} \times \frac{\mathrm{L}^{2}}{4 \pi}$ $\mathrm{L}^{2} =\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\text { So, } \quad \mathrm{L} =\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
Kerala CEE 2012
Magnetism and Matter
154039
The effective length of a magnet is $31.4 \mathrm{~cm}$ and its pole strength is $0.8 \mathrm{Am}$. The magnetic moment, if it is bent in the form of a semicircle is $\mathbf{A m}^{2}$
1 0.12
2 1.2
3 0.16
4 1.6
Explanation:
C Given, Effective length wire $(l)=31.4 \mathrm{~cm}=31.4 \times 10^{-2} \mathrm{~m}$ Pole strength $(\mathrm{m})=0.8 \mathrm{Am}$ When magnet is bent inform of semicircle - Then, circumference of semicircle $=$ length of magnet $\pi \mathrm{r}=31.4 \times 10^{-2}$ $\operatorname{Radius}(\mathrm{r})=\frac{31.4 \times 10^{-2}}{\pi}$ Diameter $=2 \mathrm{r}$ Initially magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{M}=0.8 \times 31.4 \times 10^{-2}$ $\mathrm{M}=25.12 \times 10^{-2}$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime}=0.8 \times 2 \times \frac{31.4 \times 10^{-2}}{3.14}$ $\mathrm{M}^{\prime}=0.16 \mathrm{Am}^{2}$
GUJCET 2016
Magnetism and Matter
154040
A bar magnet of magnetic moment $M$ is placed at right angles to a magnetic induction $B$. If a force $F$ is experienced by each pole of the magnet, the length of the magnet will be
1 $\mathrm{MB} / \mathrm{F}$
2 $\mathrm{BF} / \mathrm{M}$
3 $\mathrm{MF} / \mathrm{B}$
4 $\mathrm{F} / \mathrm{MB}$
Explanation:
A Given, Magnetic moment $=\mathrm{M}$ Magnetic induction $=\mathrm{B}, \theta=90^{\circ}$ We know that, $\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$ $\tau=\mathrm{MB} \sin \theta$ $\tau=\mathrm{MB} \sin 90^{\circ}$ $\tau=\mathrm{MB}$ The torque experienced by magnet- $\tau=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$ $\tau=\mathrm{rF} \sin 90^{\circ}$ $\tau=\mathrm{rF}$ Here, $\mathrm{r}=l$ ( $l=$ length of magnet) $\therefore \quad \tau=l \mathrm{~F}$ From equation (i) and (ii), we get- $l \mathrm{~F}=\mathrm{MB}$ $l=\frac{\mathrm{MB}}{\mathrm{F}}$
COMEDK 2013
Magnetism and Matter
154042
Assertion: A disc-shaped magnet is deviated above a superconducting material that has been cooled by liquid nitrogen. Reason: Super conductors repel a magnet.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A A superconductor are diamagnetic in nature. So, when it is placed above a magnet it will repelled by magnet or it will be move to from higher field to lower field. This is the principle behind deviation of superconductor above magnet. Superconductor exist only a below a certain temperature.
AIIMS-2005
Magnetism and Matter
154041
Assertion: Electromagnetic are made of soft iron. Reason: Coercivity of soft iron is small.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B As the material of electromagnet is subjected to magnetize and demagnetize, the hysteresis loss should be small. Material should gain high value I and $B$ with low value of magnetic field intensity. So the soft iron is preferred. Whose coercively is small.
154032
A straight wire carrying current $I$ is made into a circular loop. If $M$ is the magnetic moment associated with the loop, then the length of the wire is
1 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
2 $\sqrt{\frac{2 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
4 $\sqrt{\frac{\pi \mathrm{M}}{2 \mathrm{I}}}$
5 $\sqrt{\frac{\pi \mathrm{M}}{4 \mathrm{I}}}$
Explanation:
A Let, the length of wire $=\mathrm{L}$ When, wire is bent in the form of circle. Then, circumference of circle $=$ length of wire $2 \pi \mathrm{r}=\mathrm{L}$ Where, $r=$ Radius of circle Then, $r=\frac{L}{2 \pi}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=\pi \times\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}=\frac{\pi \mathrm{L}^{2}}{4 \pi^{2}}=\frac{\mathrm{L}^{2}}{4 \pi}$ We know that, Magnetic moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M} =\mathrm{I} \times \frac{\mathrm{L}^{2}}{4 \pi}$ $\mathrm{L}^{2} =\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\text { So, } \quad \mathrm{L} =\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
Kerala CEE 2012
Magnetism and Matter
154039
The effective length of a magnet is $31.4 \mathrm{~cm}$ and its pole strength is $0.8 \mathrm{Am}$. The magnetic moment, if it is bent in the form of a semicircle is $\mathbf{A m}^{2}$
1 0.12
2 1.2
3 0.16
4 1.6
Explanation:
C Given, Effective length wire $(l)=31.4 \mathrm{~cm}=31.4 \times 10^{-2} \mathrm{~m}$ Pole strength $(\mathrm{m})=0.8 \mathrm{Am}$ When magnet is bent inform of semicircle - Then, circumference of semicircle $=$ length of magnet $\pi \mathrm{r}=31.4 \times 10^{-2}$ $\operatorname{Radius}(\mathrm{r})=\frac{31.4 \times 10^{-2}}{\pi}$ Diameter $=2 \mathrm{r}$ Initially magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{M}=0.8 \times 31.4 \times 10^{-2}$ $\mathrm{M}=25.12 \times 10^{-2}$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime}=0.8 \times 2 \times \frac{31.4 \times 10^{-2}}{3.14}$ $\mathrm{M}^{\prime}=0.16 \mathrm{Am}^{2}$
GUJCET 2016
Magnetism and Matter
154040
A bar magnet of magnetic moment $M$ is placed at right angles to a magnetic induction $B$. If a force $F$ is experienced by each pole of the magnet, the length of the magnet will be
1 $\mathrm{MB} / \mathrm{F}$
2 $\mathrm{BF} / \mathrm{M}$
3 $\mathrm{MF} / \mathrm{B}$
4 $\mathrm{F} / \mathrm{MB}$
Explanation:
A Given, Magnetic moment $=\mathrm{M}$ Magnetic induction $=\mathrm{B}, \theta=90^{\circ}$ We know that, $\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$ $\tau=\mathrm{MB} \sin \theta$ $\tau=\mathrm{MB} \sin 90^{\circ}$ $\tau=\mathrm{MB}$ The torque experienced by magnet- $\tau=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$ $\tau=\mathrm{rF} \sin 90^{\circ}$ $\tau=\mathrm{rF}$ Here, $\mathrm{r}=l$ ( $l=$ length of magnet) $\therefore \quad \tau=l \mathrm{~F}$ From equation (i) and (ii), we get- $l \mathrm{~F}=\mathrm{MB}$ $l=\frac{\mathrm{MB}}{\mathrm{F}}$
COMEDK 2013
Magnetism and Matter
154042
Assertion: A disc-shaped magnet is deviated above a superconducting material that has been cooled by liquid nitrogen. Reason: Super conductors repel a magnet.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A A superconductor are diamagnetic in nature. So, when it is placed above a magnet it will repelled by magnet or it will be move to from higher field to lower field. This is the principle behind deviation of superconductor above magnet. Superconductor exist only a below a certain temperature.
AIIMS-2005
Magnetism and Matter
154041
Assertion: Electromagnetic are made of soft iron. Reason: Coercivity of soft iron is small.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B As the material of electromagnet is subjected to magnetize and demagnetize, the hysteresis loss should be small. Material should gain high value I and $B$ with low value of magnetic field intensity. So the soft iron is preferred. Whose coercively is small.
