360618
An iron rod of length \(L\) and magnetic moment \(M\) is bent in the form of a semicircle. Now its magnetic moment will be
1 \(M \pi\)
2 \(\dfrac{2 M}{\pi}\)
3 \(M\)
4 \(\dfrac{M}{\pi}\)
Explanation:
On bending a rod its pole strength remains unchanged whereas its magnetic moment changes. New magnetic moment \(M^{\prime}=m(2 R)=m\left(\dfrac{2 l}{\pi}\right)=\left(\dfrac{2 M}{\pi}\right)\)
PHXII05:MAGNETISM and MATTER
360619
A bar magnet of length \(l\) and magnetic dipole moment \(M\) is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
1 \(\dfrac{M}{2}\)
2 \(\dfrac{2}{\pi} M\)
3 \(\dfrac{3}{\pi} M\)
4 \(M\)
Explanation:
The magnetic moment, \(M=m l\left(\begin{array}{l}m=\text { pole strength } \\ l=\text { length of magnet }\end{array}\right)\) According to question, \(l=\dfrac{\pi}{3} \times r\) So, \(\quad r=\dfrac{3 l}{\pi}\) New magnetic moment, \(M^{\prime}=m \times\left(2 r \sin 30^{\circ}\right)\) \(=m \times \dfrac{3 l}{\pi}=\dfrac{3}{\pi} \cdot m l\)\(=\dfrac{3 M}{\pi}\)
PHXII05:MAGNETISM and MATTER
360620
A magnet of magnetic moment \(M\) and pole strength \(m\) is divided in two equal parts, the magnetic moment of each part will be
1 2\(M\)
2 \(M / 2\)
3 \(M\)
4 \(\sqrt{2} M\)
Explanation:
If magnet cuts parallel to length then pole strength becomes half and length will be the same \(M^{\prime}=\left(\dfrac{m}{2}\right) 2 l=M / 2\) If it cuts parallel to equatorial line than pole strength is same but length becomes half \(M^{\prime}=(m) \dfrac{2 l}{2}=\dfrac{M}{2}\)
PHXII05:MAGNETISM and MATTER
360621
The dimensional formula for magnetic moment is
1 \({M^0}\;{L^2}\;{T^0}\;{A^1}\)
2 \({M^0}\;{L^1}\;{T^0}\;{A^2}\)
3 \({M^0}{L^2}{T^0}{A^2}\)
4 \({M^0}{L^0}\;{T^1}{A^1}\)
Explanation:
Magnetic moment \( = m(2l)\) \( = (Am)m = A{m^2} = M^\circ {L^2}T^\circ A\)
360618
An iron rod of length \(L\) and magnetic moment \(M\) is bent in the form of a semicircle. Now its magnetic moment will be
1 \(M \pi\)
2 \(\dfrac{2 M}{\pi}\)
3 \(M\)
4 \(\dfrac{M}{\pi}\)
Explanation:
On bending a rod its pole strength remains unchanged whereas its magnetic moment changes. New magnetic moment \(M^{\prime}=m(2 R)=m\left(\dfrac{2 l}{\pi}\right)=\left(\dfrac{2 M}{\pi}\right)\)
PHXII05:MAGNETISM and MATTER
360619
A bar magnet of length \(l\) and magnetic dipole moment \(M\) is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
1 \(\dfrac{M}{2}\)
2 \(\dfrac{2}{\pi} M\)
3 \(\dfrac{3}{\pi} M\)
4 \(M\)
Explanation:
The magnetic moment, \(M=m l\left(\begin{array}{l}m=\text { pole strength } \\ l=\text { length of magnet }\end{array}\right)\) According to question, \(l=\dfrac{\pi}{3} \times r\) So, \(\quad r=\dfrac{3 l}{\pi}\) New magnetic moment, \(M^{\prime}=m \times\left(2 r \sin 30^{\circ}\right)\) \(=m \times \dfrac{3 l}{\pi}=\dfrac{3}{\pi} \cdot m l\)\(=\dfrac{3 M}{\pi}\)
PHXII05:MAGNETISM and MATTER
360620
A magnet of magnetic moment \(M\) and pole strength \(m\) is divided in two equal parts, the magnetic moment of each part will be
1 2\(M\)
2 \(M / 2\)
3 \(M\)
4 \(\sqrt{2} M\)
Explanation:
If magnet cuts parallel to length then pole strength becomes half and length will be the same \(M^{\prime}=\left(\dfrac{m}{2}\right) 2 l=M / 2\) If it cuts parallel to equatorial line than pole strength is same but length becomes half \(M^{\prime}=(m) \dfrac{2 l}{2}=\dfrac{M}{2}\)
PHXII05:MAGNETISM and MATTER
360621
The dimensional formula for magnetic moment is
1 \({M^0}\;{L^2}\;{T^0}\;{A^1}\)
2 \({M^0}\;{L^1}\;{T^0}\;{A^2}\)
3 \({M^0}{L^2}{T^0}{A^2}\)
4 \({M^0}{L^0}\;{T^1}{A^1}\)
Explanation:
Magnetic moment \( = m(2l)\) \( = (Am)m = A{m^2} = M^\circ {L^2}T^\circ A\)
360618
An iron rod of length \(L\) and magnetic moment \(M\) is bent in the form of a semicircle. Now its magnetic moment will be
1 \(M \pi\)
2 \(\dfrac{2 M}{\pi}\)
3 \(M\)
4 \(\dfrac{M}{\pi}\)
Explanation:
On bending a rod its pole strength remains unchanged whereas its magnetic moment changes. New magnetic moment \(M^{\prime}=m(2 R)=m\left(\dfrac{2 l}{\pi}\right)=\left(\dfrac{2 M}{\pi}\right)\)
PHXII05:MAGNETISM and MATTER
360619
A bar magnet of length \(l\) and magnetic dipole moment \(M\) is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
1 \(\dfrac{M}{2}\)
2 \(\dfrac{2}{\pi} M\)
3 \(\dfrac{3}{\pi} M\)
4 \(M\)
Explanation:
The magnetic moment, \(M=m l\left(\begin{array}{l}m=\text { pole strength } \\ l=\text { length of magnet }\end{array}\right)\) According to question, \(l=\dfrac{\pi}{3} \times r\) So, \(\quad r=\dfrac{3 l}{\pi}\) New magnetic moment, \(M^{\prime}=m \times\left(2 r \sin 30^{\circ}\right)\) \(=m \times \dfrac{3 l}{\pi}=\dfrac{3}{\pi} \cdot m l\)\(=\dfrac{3 M}{\pi}\)
PHXII05:MAGNETISM and MATTER
360620
A magnet of magnetic moment \(M\) and pole strength \(m\) is divided in two equal parts, the magnetic moment of each part will be
1 2\(M\)
2 \(M / 2\)
3 \(M\)
4 \(\sqrt{2} M\)
Explanation:
If magnet cuts parallel to length then pole strength becomes half and length will be the same \(M^{\prime}=\left(\dfrac{m}{2}\right) 2 l=M / 2\) If it cuts parallel to equatorial line than pole strength is same but length becomes half \(M^{\prime}=(m) \dfrac{2 l}{2}=\dfrac{M}{2}\)
PHXII05:MAGNETISM and MATTER
360621
The dimensional formula for magnetic moment is
1 \({M^0}\;{L^2}\;{T^0}\;{A^1}\)
2 \({M^0}\;{L^1}\;{T^0}\;{A^2}\)
3 \({M^0}{L^2}{T^0}{A^2}\)
4 \({M^0}{L^0}\;{T^1}{A^1}\)
Explanation:
Magnetic moment \( = m(2l)\) \( = (Am)m = A{m^2} = M^\circ {L^2}T^\circ A\)
360618
An iron rod of length \(L\) and magnetic moment \(M\) is bent in the form of a semicircle. Now its magnetic moment will be
1 \(M \pi\)
2 \(\dfrac{2 M}{\pi}\)
3 \(M\)
4 \(\dfrac{M}{\pi}\)
Explanation:
On bending a rod its pole strength remains unchanged whereas its magnetic moment changes. New magnetic moment \(M^{\prime}=m(2 R)=m\left(\dfrac{2 l}{\pi}\right)=\left(\dfrac{2 M}{\pi}\right)\)
PHXII05:MAGNETISM and MATTER
360619
A bar magnet of length \(l\) and magnetic dipole moment \(M\) is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
1 \(\dfrac{M}{2}\)
2 \(\dfrac{2}{\pi} M\)
3 \(\dfrac{3}{\pi} M\)
4 \(M\)
Explanation:
The magnetic moment, \(M=m l\left(\begin{array}{l}m=\text { pole strength } \\ l=\text { length of magnet }\end{array}\right)\) According to question, \(l=\dfrac{\pi}{3} \times r\) So, \(\quad r=\dfrac{3 l}{\pi}\) New magnetic moment, \(M^{\prime}=m \times\left(2 r \sin 30^{\circ}\right)\) \(=m \times \dfrac{3 l}{\pi}=\dfrac{3}{\pi} \cdot m l\)\(=\dfrac{3 M}{\pi}\)
PHXII05:MAGNETISM and MATTER
360620
A magnet of magnetic moment \(M\) and pole strength \(m\) is divided in two equal parts, the magnetic moment of each part will be
1 2\(M\)
2 \(M / 2\)
3 \(M\)
4 \(\sqrt{2} M\)
Explanation:
If magnet cuts parallel to length then pole strength becomes half and length will be the same \(M^{\prime}=\left(\dfrac{m}{2}\right) 2 l=M / 2\) If it cuts parallel to equatorial line than pole strength is same but length becomes half \(M^{\prime}=(m) \dfrac{2 l}{2}=\dfrac{M}{2}\)
PHXII05:MAGNETISM and MATTER
360621
The dimensional formula for magnetic moment is
1 \({M^0}\;{L^2}\;{T^0}\;{A^1}\)
2 \({M^0}\;{L^1}\;{T^0}\;{A^2}\)
3 \({M^0}{L^2}{T^0}{A^2}\)
4 \({M^0}{L^0}\;{T^1}{A^1}\)
Explanation:
Magnetic moment \( = m(2l)\) \( = (Am)m = A{m^2} = M^\circ {L^2}T^\circ A\)
